Question 26 — JEE Main 24 Jan 2026 (S I)

Question

Let a line L passing through the point $\mathrm{P}(1,1,1)$ be perpendicular to the
lines $\frac{x-4}{4}=\frac{y-1}{1}=\frac{z-1}{1}$ and $\frac{x-17}{1}=\frac{y-71}{1}=\frac{z}{0}$. Let the line L
intersect the yz -plane at the point Q . Another line parallel to L and passing through the point $\mathrm{S}(1,0,-1)$ intersects the yz -plane at the point R . Then the square of the area of the parallelogram PQRS is equal to $\_\_\_\_$ .

[JEE MAIN 24 Jan 2026 S I]

Solution

$\mathrm{L}_{1} \xrightarrow{\text { drs }} 4,1,1, \mathrm{~L}_{2} \xrightarrow{\text { drs }} 1,1,0$
required $\mathrm{L}, \mathrm{L} \perp \mathrm{L}_{1}$ \& $\mathrm{L} \perp \mathrm{L}_{2}$
$\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 4 & 1 & 1 \\ 1 & 1 & 0\end{array}\right|=-\hat{i}+\hat{j}+3 \hat{j}$
line $\mathrm{L} \xrightarrow{\text { drs }}-1,1,3$
line $\mathrm{L} \frac{\mathrm{x}-1}{-1}=\frac{\mathrm{y}-1}{1}=\frac{\mathrm{z}-1}{3}=\lambda \mathrm{Q}(-\lambda+1, \lambda+1,3 \lambda+1)$
Q lies on yz plane
$\mathrm{Q}_{\mathrm{x}}=0 \Rightarrow-\lambda+1=0 \Rightarrow \lambda=1$
$\mathrm{Q}(0,2,4) \mathrm{S}(1,0,-1)$
$\overrightarrow{\mathrm{PQ}}=-\hat{i}+\hat{j}+3 \hat{k}, \overrightarrow{\mathrm{PS}}=-\hat{j}-2 \hat{k}$

area of parallelogram $=|\overrightarrow{\mathrm{PQ}} \times \overrightarrow{\mathrm{PS}}|=\left|\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ -1 & 1 & 3 \\ 0 & 1 & -2\end{array}\right|$