MHT CET 2025 23rd April Morning Shift (Mathematics) Question Paper with Solutions

$\begin{array}{cc}& p=\text{ Probability of getting head }=\frac{1}{2}\\ & \therefore q=1-p=1-\frac{1}{2}=\frac{1}{2}\\ & \therefore \begin{array}{c}\text{ Also, }n=100\\ \text{ Required probability }\\ =P(X=0)+P(X=2)+P(X=4)+\dots +P(X=100)\\ ={}^{100}{C}_0{\left(\frac{1}{2}\right)}^0{\left(\frac{1}{2}\right)}^{100}+{}^{100}{C}_2{\left(\frac{1}{2}\right)}^2{\left(\frac{1}{2}\right)}^{98}\end{array}\\ & +{}^{100}{C}_4{\left(\frac{1}{2}\right)}^4{\left(\frac{1}{2}\right)}^{96}+\dots +{}^{100}{C}_{100}{\left(\frac{1}{2}\right)}^{100}{\left(\frac{1}{2}\right)}^0\\ & ={\left(\frac{1}{2}\right)}^{100}({}^{100}{C}_0+{}^{100}{C}_2+{}^{100}{C}_4+\dots +{}^{100}{C}_{100})\\ & ={\left(\frac{1}{2}\right)}^{100}\left(2^{100-1}\right)\\ & =\frac{1}{2^{100}}\left(2^{99}\right)\\ & =\frac{1}{2}\end{array}$

$\begin{array}{cc}& \begin{array}{cc}& \mid \begin{array}{cc}\overline{𝑎}\cdot \overline{𝑐}& \overline{𝑏}\cdot \overline{𝑐}\\ \overline{𝑎}\cdot \overline{𝑑}& \overline{𝑏}\cdot \overline{𝑑}\end{array}\mid =0\\ & \Rightarrow (\overline{𝑎}\cdot \overline{𝑐})(\overline{𝑏}\cdot \overline{𝑑})-(\overline{𝑎}\cdot \overline{𝑑})(\overline{𝑏}\cdot \overline{𝑐})=0\end{array}\\ & \text{ If }\bar{a},\bar{b},\bar{c},\bar{d}\text{ are four vectors, then }\\ & (\bar{a}\times \bar{b})\cdot (\bar{c}\times \bar{d})=(\bar{a}\cdot \bar{c})(\bar{b}\cdot \bar{d})-(\bar{a}\cdot \bar{d})(\bar{b}\cdot \bar{c})=0\end{array}$

$\begin{array}{cc}& \Rightarrow (2\hat{i}+3\hat{j}-\hat{k})\cdot (3\hat{i}+2\hat{j}+𝜆\hat{k})=0\\ & \Rightarrow 2\left(3\right)+3\left(2\right)+(-1)\left(𝜆\right)=0\\ & \Rightarrow 𝜆=12\end{array}$

$\begin{array}{cc}& \cos \frac{𝐵}{2}=\sqrt{\surd \frac{𝑠(𝑠-𝑏)}{𝑐𝑎}}\dots \left[\text{ By half angle formula }\right]\\ & \Rightarrow \sqrt{\surd \frac{𝑐+𝑎}{2𝑎}}=\sqrt{\surd \frac{𝑠(𝑠-𝑏)}{𝑐𝑎}}\\ & \Rightarrow \frac{𝑐+𝑎}{2𝑎}=\frac{𝑠(𝑠-𝑏)}{𝑐𝑎}\\ & \Rightarrow \frac{𝑐+𝑎}{2𝑎}=\frac{\frac{𝑎+𝑏+𝑐}{2}(\frac{𝑎+𝑏+𝑐}{2}-𝑏)}{𝑐𝑎}\\ & \Rightarrow \frac{𝑐+𝑎}{2𝑎}=\frac{(𝑎+𝑐+𝑏)(𝑎+𝑐-𝑏)}{4𝑎𝑐}\\ & \Rightarrow \frac{𝑐+𝑎}{2𝑎}=\frac{(𝑎+𝑐{)}^2-{𝑏}^2}{4𝑎𝑐}\\ & \Rightarrow \frac{𝑐+𝑎}{2𝑎}=\frac{{𝑎}^2+2𝑎𝑐+{𝑐}^2-{𝑏}^2}{4𝑎𝑐}\\ & \Rightarrow 2{𝑐}^2+2𝑎𝑐={𝑎}^2+2𝑎𝑐+{𝑐}^2-{𝑏}^2\\ & \Rightarrow {𝑎}^2={𝑏}^2+{𝑐}^2\end{array}$

The equations of the bisectors of the co-ordinate axes are $𝑥-𝑦=0$ and $𝑥+𝑦=0$.

Slope of $𝑥-𝑦=0$ is 1 .

Slope of $𝑥+𝑦=0$ is -1 .

$\therefore$ Equation of the line passing through $(-2,3)$ and having slope 1 is

$\begin{array}{cc}& 𝑦-3=1(𝑥+2)\\ & \Rightarrow 𝑥-𝑦+5=0\end{array}$

Equation of the line passing through $(-2,3)$ and having slope -1 is

$\begin{array}{cc}& 𝑦-3=-1(𝑥+2)\\ & \Rightarrow 𝑥+𝑦-1=0\end{array}$

$\therefore$ The joint equation is

$\begin{array}{cc}& (𝑥-𝑦+5)(𝑥+𝑦-1)=0\\ & \Rightarrow {𝑥}^2-{𝑦}^2+4𝑥+6𝑦-5=0\end{array}$

$\begin{array}{cc}\text{ Radius of circle }& =\mathrm{OA}\\ & =\sqrt{\surd (2-0{)}^2+(4-0{)}^2}\\ & =\sqrt{\surd 4+16}\\ & =2\sqrt{\surd 5}\end{array}$

The length of the side of an equilateral triangle inscribed in the circle ${𝑥}^2+{𝑦}^2=a^2$ is a $\sqrt{\surd 3}$.

$\therefore$ Length of side $=\left(2\sqrt{\surd 5}\right)\left(\sqrt{\surd 3}\right)=2\sqrt{\surd 15}$

In $△\mathrm{AMC}$,

$\cos {30}^{\circ }=\frac{\mathrm{AM}}{\mathrm{AC}}$

$\Rightarrow \frac{\sqrt{\surd 3}}{2}=\frac{\mathrm{AM}}{2\sqrt{\surd 15}}$

$\Rightarrow \mathrm{AM}=3\sqrt{\surd 5}$ units

$\begin{array}{cc}& (\overline{𝑎}\times \overline{𝑏})=\overline{𝑐}\\ & \Rightarrow (\overline{𝑎}\times \overline{𝑏})\cdot \overline{𝑐}=\overline{𝑐}\cdot \overline{𝑐}\\ & =(\widehat{𝑗}-\widehat{𝑘})\cdot (\widehat{𝑗}-\widehat{𝑘})\\ & \therefore (\overline{𝑎}\times \overline{𝑏})\cdot \overline{𝑐}=2\\ & \Rightarrow \overline{𝑎}\cdot (\overline{𝑏}\times \overline{𝑐})=2\\ & \Rightarrow \overline{𝑏}\cdot (\overline{𝑐}\times \overline{𝑎})=2\dots [\because \overline{𝑎}\cdot (\overline{𝑏}\times \overline{𝑐})=\overline{𝑏}\cdot (\overline{𝑐}\times \overline{𝑎}\left)\right]\\ & \Rightarrow \overline{𝑏}\cdot (\overline{𝑎}\times \overline{𝑐})=-2\\ & \overline{𝑎}\times \overline{𝑐}=\mid \begin{array}{ccc}\widehat{𝑖}& \widehat{𝑗}& \widehat{𝑘}\\ 1& 1& 1\\ 0& 1& -1\end{array}\mid \\ & =-2\widehat{𝑖}+\widehat{𝑗}+\widehat{𝑘}\\ & |\overline{𝑎}\times \overline{𝑐}|=\sqrt{\surd 4+1+1}\\ & =\sqrt{\surd 6}\end{array}$

Length of projection vector of $\overline{𝑏}$ on $\overline{𝑎}\times \overline{𝑐}$

$=\frac{|\bar{b}\cdot (\bar{a}\times \bar{c}\left)\right|}{|\bar{a}\times \bar{c}|}$

$\Rightarrow 𝑙=\frac{2}{\sqrt{\surd 6}}$

$\Rightarrow {𝑙}^2=\frac{4}{6}$

$\Rightarrow 3{𝑙}^2=2$

Let the equation of the line L parallel to $3𝑥-𝑦-2=0$ be $3𝑥-𝑦+𝜆=0$.

Since it passes through $(1,2)$.

$\begin{array}{cc}\therefore & 3\left(1\right)-2+𝜆=0\\ & \Rightarrow 𝜆=-1\end{array}$

$\therefore$ Equation of the line L is $3𝑥-𝑦-1=0$.

The point of intersection of $3𝑥-𝑦-1=0$ and $𝑥+𝑦=0$ is $(\frac{1}{4},\frac{-1}{4})$.

$\begin{array}{cc}& \therefore \text{ Distance between }(1,2)\text{ and }(\frac{1}{4},\frac{-1}{4})\text{ is }\\ & \begin{array}{cc}& \sqrt{\surd {(1-\frac{1}{4})}^2+{(2+\frac{1}{4})}^2}\\ =& \sqrt{\surd \frac{9}{16}+\frac{81}{16}}\\ =& \sqrt{\surd \frac{90}{16}}=\frac{3\sqrt{\surd 10}}{4}\text{ units }\end{array}\end{array}$

$\begin{array}{cc}& \underset{𝑥\to 2}{lim}\frac{𝑥+3{𝑥}^2+5{𝑥}^3+7{𝑥}^4-166}{𝑥-2}\\ & =\underset{𝑥\to 2}{lim}\frac{(𝑥-2)(7{𝑥}^3+19{𝑥}^2+41𝑥+83)}{𝑥-2}\\ & =\underset{𝑥\to 2}{lim}(7{𝑥}^3+19{𝑥}^2+41𝑥+83)\\ & =7(2{)}^3+19(2{)}^2+41\left(2\right)+83\\ & =56+76+82+83\\ & =297\end{array}$

$\begin{array}{cc}& \text{ Since }f\left(𝑥\right)\text{ is continuous at }𝑥=0\text{. }\\ & \begin{array}{cc}\therefore 𝑓\left(0\right)& =\underset{𝑥\to 0}{lim}\frac{{10}^{𝑥}+7^{𝑥}-{14}^{𝑥}-5^{𝑥}}{1-\cos 𝑥}\\ & =\underset{𝑥\to 0}{lim}\frac{2^{𝑥}\cdot 5^{𝑥}-5^{𝑥}-2^{𝑥}\cdot 7^{𝑥}+7^{𝑥}}{1-\cos 𝑥}\\ & =\underset{𝑥\to 0}{lim}\frac{(2^{𝑥}-1)(5^{𝑥}-7^{𝑥})}{1-\cos 𝑥}\\ & =\underset{𝑥\to 0}{lim}\frac{\left(\frac{2^{𝑥}-1}{𝑥}\right)(\frac{5^{𝑥}-1}{𝑥}-\frac{7^{𝑥}-1}{𝑥})}{\frac{1-\cos 𝑥}{{𝑥}^2}}\\ & =\frac{(\log 2)(\log 5-\log 7)}{\frac{1}{2}}\\ & =2\log 2\log \left(\frac{5}{7}\right)\\ & =\log 4\log \left(\frac{5}{7}\right)\end{array}\end{array}$

$\begin{array}{cc}& \begin{array}{cc}& (\mathrm{AB}{)}^{-1}=\frac{1}{6}\left[\begin{array}{cc}-7& -3\\ 2& 3\end{array}\right]\\ & \Rightarrow B^{-1}{A}^{-1}=\frac{1}{6}\left[\begin{array}{cc}-7& -3\\ 2& 3\end{array}\right]\\ & \Rightarrow B^{-1}\left(\frac{1}{3}\left[\begin{array}{cc}4& 3\\ -1& 0\end{array}\right]\right)=\frac{1}{6}\left[\begin{array}{cc}-7& -3\\ 2& 3\end{array}\right]\end{array}\\ & \text{ Only option (C) satisfies the above condition. }\end{array}$

$\begin{array}{cc}& \frac{\sin A+\sin B}{\cos A+\cos B}=\frac{𝑥}{𝑦}\\ & \Rightarrow \frac{2\sin \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right)}{2\cos \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right)}=\frac{𝑥}{𝑦}\\ & \Rightarrow \tan \left(\frac{A+B}{2}\right)=\frac{𝑥}{𝑦}\\ & \begin{array}{cc}\sin (A+B)& =\frac{2\tan \left(\frac{A+B}{2}\right)}{1+{\tan }^2\left(\frac{A+B}{2}\right)}\\ & =\frac{2\left(\frac{𝑥}{𝑦}\right)}{1+{\left(\frac{𝑥}{𝑦}\right)}^2}\\ & =\frac{2𝑥𝑦}{{𝑥}^2+{𝑦}^2}\end{array}\end{array}$

The sum of all the probabilities in a probability distribution is always unity.

$\begin{array}{cc}& \therefore \frac{1}{4}+𝐾+\frac{1}{4}+\frac{1}{3}=1\\ & \Rightarrow 𝐾+\frac{5}{6}=1\\ & \Rightarrow 𝐾=\frac{1}{6}\\ & \text{ Mean }\left(𝜇\right)=-3\left(\frac{1}{4}\right)+0+1\left(\frac{1}{4}\right)+𝛼\left(\frac{1}{3}\right)\\ & \Rightarrow 𝜇=\frac{𝛼}{3}-\frac{1}{2}...\left(𝑖\right)\\ & \begin{array}{cc}{𝜎}^2& =(-3{)}^2\left(\frac{1}{4}\right)+0+1^2\left(\frac{1}{4}\right)+{𝛼}^2\left(\frac{1}{3}\right)-{𝜇}^2\\ & =\frac{10}{4}+\frac{{𝛼}^2}{3}-{(\frac{𝛼}{3}-\frac{1}{2})}^2\\ & =\frac{5}{2}+\frac{{𝛼}^2}{3}-\frac{{𝛼}^2}{9}+\frac{𝛼}{3}-\frac{1}{4}\end{array}\end{array}$

$\begin{array}{cc}\therefore & {𝜎}^2=\frac{2{𝛼}^2}{9}+\frac{𝛼}{3}+\frac{9}{4}...\left(𝑖𝑖\right)\\ & 𝜎-𝜇=2\dots \left[𝐺𝑖𝑣𝑒𝑛\right]\\ & \Rightarrow 𝜎=2+𝜇\\ & \Rightarrow 𝜎=\frac{𝛼}{3}+\frac{3}{2}\dots \left[𝐹𝑟𝑜𝑚\right(𝑖\left)\right]\\ & \Rightarrow {𝜎}^2=\frac{{𝛼}^2}{9}+𝛼+\frac{9}{4}\\ & \Rightarrow \frac{2{𝛼}^2}{9}+\frac{𝛼}{3}+\frac{9}{4}=\frac{{𝛼}^2}{9}+𝛼+\frac{9}{4}\dots \left[𝐹𝑟𝑜𝑚\right(𝑖𝑖\left)\right]\\ & \Rightarrow \frac{{𝛼}^2}{9}-\frac{2𝛼}{3}=0\end{array}$

$\begin{array}{cc}& \begin{array}{cc}& \Rightarrow {𝛼}^2-6𝛼=0\\ & \Rightarrow 𝛼(𝛼-6)=0\\ & \Rightarrow 𝛼=0\text{ or }𝛼=6\\ & \Rightarrow 𝛼=6\\ & 𝜇=\frac{6}{3}-\frac{1}{2}...\left[𝐹𝑟𝑜𝑚\right(𝑖\left)\right]\\ & =\frac{3}{2}\\ & 𝜎-𝜇=2\\ & \Rightarrow 𝜎-\frac{3}{2}=2\\ & \Rightarrow 𝜎=\frac{7}{2}\end{array}\\ & \end{array}$

$\begin{array}{cc}& 𝑃\left(\text{ getting }5\text{ or }6\right)=\frac{2}{6}=\frac{1}{3}\\ & 𝑃\left(\text{ not getting }5\text{ or }6\right)=1-\frac{1}{3}=\frac{2}{3}\end{array}$

Let X be the net gain or loss in rupees.

Case I: Wins on the first throw.

$\begin{array}{cc}& 𝑋=100\\ \therefore & 𝑃(𝑋=100)=\frac{1}{3}\end{array}$

Case II: Loses the first, wins on the second throw.

$\begin{array}{cc}& 𝑋=-50+100=50\\ \therefore & 𝑃(𝑋=50)=\frac{2}{3}\times \frac{1}{3}=\frac{2}{9}\end{array}$

Case III: Loses the first two, wins on the third throw.

$\begin{array}{cc}& 𝑋=-50-50+100=0\\ \therefore & 𝑃(𝑋=0)=\frac{2}{3}\times \frac{2}{3}\times \frac{1}{3}=\frac{4}{27}\end{array}$

Case IV: Loses all three throws.

$\begin{array}{cc}𝑋=& -50-50-50=-150\\ \therefore 𝑃(𝑋=& -150)=\frac{2}{3}\times \frac{2}{3}\times \frac{2}{3}=\frac{8}{27}\\ 𝐸\left(𝑋\right)& =100\left(\frac{1}{3}\right)+50\left(\frac{2}{9}\right)+0+(-150)\left(\frac{8}{27}\right)\\ & =\frac{900+300-1200}{27}\\ & =0\end{array}$

$\therefore$ The expected gain/loss is $𝑅𝑠0$.

Let $T_0$ be the air temperature and T be temperature of the substance at time $𝑡$. Then, $\frac{\mathrm{dT}}{\mathrm{dt}}=-k(T-T_0)$, where $k>0$

$\begin{array}{cc}& \Rightarrow \frac{\mathrm{dT}}{\mathrm{dt}}=-k(T-290)\\ & \Rightarrow \frac{\mathrm{dT}}{T-290}=-\mathrm{kdt}\end{array}$

$\begin{array}{cc}& \text{ Integrating on both sides, we get }\\ & \begin{array}{cc}& \log (T-290)=-\mathrm{kt}+c\\ & \text{ When }t=0,T=370\\ & \therefore \log (370-290)=0+c\\ & \Rightarrow c=\log 80\\ & \therefore \log (T-290)=-\mathrm{kt}+\log 80...\left(𝑖\right)\end{array}\end{array}$

$\begin{array}{cc}& \text{ When }t=10,T=330\\ & \therefore \log (330-290)=-10k+\log 80\\ & \Rightarrow \log 40=-10k+\log 80\\ & \Rightarrow 10k=\log \frac{80}{40}\\ & \Rightarrow k=\frac{1}{10}\log 2\\ & \therefore \log (T-290)=\frac{-t}{10}\log 2+\log 80\end{array}$

                                                                              …[From (i)]

When $𝑇=295$, we have

$\begin{array}{cc}& \log (295-290)=\frac{-𝑡}{10}\log 2+\log 80\\ & \Rightarrow \frac{𝑡}{10}\log 2=\log \frac{80}{5}\\ & \Rightarrow \frac{𝑡}{10}\log 2=\log 16\\ & \Rightarrow 𝑡=10\frac{\log 16}{\log 2}\\ & =10\frac{\log 2^4}{\log 2}=40\min \end{array}$

$\begin{array}{cc}& \begin{array}{cc}& 𝑥\frac{d𝑦}{d𝑥}=𝑦(\log 𝑦-\log 𝑥+1)\\ & \Rightarrow \frac{d𝑦}{d𝑥}=\frac{𝑦}{𝑥}[\log \left(\frac{𝑦}{𝑥}\right)+1]\\ & \text{ Put }v=\frac{𝑦}{𝑥}\\ \therefore & 𝑦=v𝑥\\ \therefore & \frac{d𝑦}{d𝑥}=v+𝑥\frac{\mathrm{dv}}{d𝑥}\end{array}\\ & \text{ From (i), }\\ & \begin{array}{cc}& v+𝑥\frac{\mathrm{dv}}{\mathrm{dx}}=v(\log v+1)\\ & \Rightarrow v+𝑥\frac{\mathrm{dv}}{\mathrm{dx}}=v\log v+v\\ & \Rightarrow 𝑥\frac{\mathrm{dv}}{\mathrm{dx}}=v\log v\\ & \Rightarrow \frac{1}{v\log v}\mathrm{dv}=\frac{d𝑥}{𝑥}\end{array}\end{array}$

$\begin{array}{cc}& \text{ Integrating on both sides, we get }\\ & \begin{array}{cc}& \int \frac{1}{𝑣\log 𝑣}𝑑𝑣=\int \frac{𝑑𝑥}{𝑥}\\ & \Rightarrow \log (\log 𝑣)=\log 𝑥+\log 𝑐\\ & \Rightarrow \log (\log 𝑣)=\log 𝑥+\log 𝑐\\ & \Rightarrow \log (\log 𝑣)=\log \left(𝑥𝑐\right)\\ & \Rightarrow \log 𝑣=𝑥𝑐\\ & \Rightarrow \log \left(\frac{𝑦}{𝑥}\right)=𝑐𝑥\end{array}\end{array}$

$\begin{array}{cc}& \sqrt{\surd \frac{d𝑦}{d𝑥}}-4\frac{d𝑦}{d𝑥}-7𝑥=0\\ & \Rightarrow {⎛⎝\sqrt{\surd \frac{d𝑦}{d𝑥}}⎞⎠}^2={(4\frac{d𝑦}{d𝑥}+7𝑥)}^2\\ & \Rightarrow \frac{d𝑦}{d𝑥}=16{\left(\frac{d𝑦}{d𝑥}\right)}^2+56𝑥\frac{d𝑦}{d𝑥}+49{𝑥}^2\end{array}$

This is a differential equation of order 1 and degree 2 .

Since the centres of all circles lie on the X-axis.

Let $(h,0)$ be the centre of the circle.

Radius of the circle $=h$

Then, the equation of the circle is

$\begin{array}{cc}& (𝑥-h{)}^2+(𝑦-0{)}^2=h^2\\ & \therefore {𝑥}^2-2𝑥h+h^2+{𝑦}^2=h^2\\ & \therefore {𝑥}^2-2𝑥h+{𝑦}^2=0...\left(𝑖\right)\end{array}$

Differentiating w.r.t. $𝑥$, we get

$\begin{array}{cc}& 2𝑥-2h+2𝑦\frac{d𝑦}{d𝑥}=0\\ \therefore & h=𝑥+𝑦\frac{d𝑦}{d𝑥}...\left(𝑖𝑖\right)\end{array}$

Substituting (ii) in (i), we get

$\begin{array}{cc}& {𝑥}^2-2𝑥(𝑥+𝑦\frac{d𝑦}{d𝑥})+{𝑦}^2=0\\ \therefore & {𝑥}^2-2{𝑥}^2-2𝑥𝑦\frac{d𝑦}{d𝑥}+{𝑦}^2=0\\ \therefore & {𝑥}^2-{𝑦}^2+2𝑥𝑦\frac{d𝑦}{d𝑥}=0\end{array}$

The points of intersection of ${𝑥}^2=8𝑦$ and $𝑥-8𝑦+2=0$ are $(-1,\frac{1}{8})$ and $(2,\frac{1}{2})$.

$\begin{array}{cc}\text{ Required area }& ={\int }_2^{-1}\left(\frac{2+𝑥}{8}\right)d𝑥-{\int }_2^{-1}\frac{{𝑥}^2}{8}d𝑥\\ & =\frac{1}{8}{[2𝑥+\frac{{𝑥}^2}{2}]}_2^{-1}-\frac{1}{8}{\left[\frac{{𝑥}^3}{3}\right]}_2^{-1}\\ & =\frac{1}{8}\left(\frac{15}{2}\right)-\frac{1}{8}\left(3\right)\\ & =\frac{9}{16}\text{ sq. units }\end{array}$

Using half angle formula,

$\begin{array}{cc}\tan \frac{𝐴}{2}& =\sqrt{\surd \frac{(𝑠-𝑏)(𝑠-𝑐)}{𝑠(𝑠-𝑎)}}\\ & =\sqrt{\surd \frac{𝑠(𝑠-𝑎)(𝑠-𝑏)(𝑠-𝑐)}{{𝑠}^2(𝑠-𝑎{)}^2}}\end{array}$

$\therefore \tan \frac{A}{2}=\frac{\Delta }{s(s-a)}$, where $\Delta$ is area of triangle ABC

Similarly,

$\begin{array}{cc}& \tan \frac{𝐵}{2}=\frac{\Delta }{𝑠(𝑠-𝑏)},\tan \frac{𝐶}{2}=\frac{\Delta }{𝑠(𝑠-𝑐)}\\ & \cot \frac{𝐴}{2}+\cot \frac{𝐵}{2}+\cot \frac{𝐶}{2}\\ & =\frac{1}{\tan \frac{𝐴}{2}}+\frac{1}{\tan \frac{𝐵}{2}}+\frac{1}{\tan \frac{𝐶}{2}}\\ & =\frac{𝑠(𝑠-𝑎)}{\Delta }+\frac{𝑠(𝑠-𝑏)}{\Delta }+\frac{𝑠(𝑠-𝑐)}{\Delta }\\ & =\frac{𝑠}{\Delta }[3𝑠-(𝑎+𝑏+𝑐\left)\right]\\ & =\frac{𝑠}{\Delta }(3𝑠-2𝑠)=\frac{{𝑠}^2}{\Delta }\end{array}$

$\begin{array}{cc}& \overline{\mathrm{PQ}}=(\hat{i}+\hat{j}+\hat{k})-(-\hat{i}+\hat{j}+\hat{k})\\ & =2\hat{i}\\ & \overline{\mathrm{QR}}=(\hat{i}-\hat{j}+\hat{k})-(\hat{i}+\hat{j}+\hat{k})\\ & =-2\hat{j}\\ & \therefore \left|\overline{\mathrm{PQ}}\right|=2\text{ and }\left|\overline{\mathrm{QR}}\right|=2\\ & \therefore \text{ Area of rectangle }=2\times 2=4\text{ sq. units }\end{array}$