JEE Main 2026 — 8 April, Evening Shift — Question 40

$\mathrm{L}_{2}: \frac{\mathrm{x}-3}{1}=\frac{\mathrm{y}+1}{2}=\frac{\mathrm{z}-4}{2}=\mathrm{t}$ $\mathrm{L}_{5}: \frac{\mathrm{x}-3}{2}=\frac{\mathrm{y}-3}{2}=\frac{\mathrm{z}-2}{1}=\mathrm{s}$ vector $\perp^{\text {ar }}$ to $\mathrm{L}_{2} \& \mathrm{~L}_{3}$ $\Rightarrow\left|\begin{array}{lll}\mathrm{i} & \mathrm{j} & \mathrm{k} \\ 1 & 2 & 2\\ 2 & 2 & 1\end{array}\right|=\hat{\mathrm{i}}(-2)-\hat{\mathrm{j}}(-3)+\hat{\mathrm{k}}(-2)$ $\Rightarrow-2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}-2 \hat{\mathrm{k}}$ $\Rightarrow-(2 \hat{i}-3 \hat{j}+2 \hat{k})$ Now, $\mathrm{L}_{1}: \frac{\mathrm{x}}{2}=\frac{\mathrm{y}}{-3}=\frac{\mathrm{z}}{2}=\ell$ intersection of $\mathrm{L}_{1}$ & $\mathrm{L}_{2}$ $\Rightarrow \mathrm{t}+3=2 \ell, 2 \mathrm{t}-1=-3 \ell, 2 \mathrm{t}+4=2 \ell$ $\mathrm{t}+3=2 \mathrm{t}+4$ $\mathrm{t}=-1$ $\Rightarrow \mathrm{P} \equiv(2,-3,2)$ A point on $\mathrm{L}_{3}$ is $\mathrm{Q}(2 \mathrm{~s}+3,2 \mathrm{~s}+3, \mathrm{~s}+2)$ $\Rightarrow \mathrm{PQ}^{2}=(2 \mathrm{~s}+1)^{2}+(2 \mathrm{~s}+6)^{2}+\mathrm{s}^{2}=17$ $\Rightarrow 9 \mathrm{~s}^{2}+28 \mathrm{~s}+20=0$ $\Rightarrow(9 \mathrm{~s}+10)(\mathrm{s}+2)=0 \Rightarrow \mathrm{~s}=-2$ $\therefore \mathrm{Q} \equiv(-1,-1,0) \equiv(\mathrm{a}, \mathrm{b}, \mathrm{c}) \Rightarrow(\mathrm{a}+\mathrm{b}+\mathrm{c})^{2}=4$