JEE Main 2025 — 7 April, Evening Shift — Question 37

$P(5,1,-3)$

$L_{1}: x-1=y-2=z=\lambda$

$L_{2}: x-2=y=z-1=\mu$

Any point of $L_{1}$ is $Q(\lambda+1, \lambda+2, \lambda)$

Any point of $L_{2}$ is $R(\mu+2, \mu, \mu+1)$

Now $P Q<\lambda-4, \lambda+1, \lambda+3>\cdot<1,1,1>=0$

$\lambda-4+\lambda+1+\lambda+3=0$

$3 \lambda=0$

$\Rightarrow \lambda=0$

$\therefore Q(1,2,0)$

Also, $P R<\mu-3, \mu-1, \mu+4>\cdot<1,1,1>=0$

$\mu-3+\mu-1+\mu+4=0$

$\Rightarrow \mu=0$

$R(2,0,1)$

Area $=\frac{1}{2}|\overrightarrow{P Q} \times \overrightarrow{P R}|=\left|\frac{1}{2}\right| \left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k}\\ 4 & -1 & -3\\ 3 & 1 & -4\end{array}\right|$

$=\frac{1}{2}|7 \hat{i}+7 \hat{j}+7 \hat{k}|=\frac{1}{2} \times 7 \times \sqrt{3}=\frac{7 \sqrt{3}}{2}$

$\because A=\frac{7 \sqrt{3}}{2}$

$$\begin{aligned} \Rightarrow 4 A^{2} & =(7 \sqrt{3})^{2} \\& =49 \times 3 \\& =147 \end{aligned}$$