[shahkb4]

Let a line L passing through the point $\mathrm{P}(1,1,1)$ be perpendicular to the lines $\frac{\mathrm{x}-4}{4}=\frac{\mathrm{y}-1}{1}=\frac{\mathrm{z}-1}{1}$ and $\frac{\mathrm{x}-17}{1}=\frac{\mathrm{y}-71}{1}=\frac{\mathrm{z}}{0}$. Let the line L intersect the yz -plane at the point Q . Another line parallel to L and passing through the point 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]

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[shahkb4]

Solution

\[ \mathrm{L}_{1} \xrightarrow{\mathrm{drs}} 4,1,1, \mathrm{~L}_{2} \xrightarrow{\mathrm{drs}} 1,1,0 \]

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 $\mathrm{L} \xrightarrow{\mathrm{drs}}-1,1,3$
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

\[ \begin{aligned} & \mathrm{Q}_{\mathrm{x}}=0 \Rightarrow-\lambda+1=0 \Rightarrow \lambda=1 \\ & \mathrm{Q}(0,2,4) \mathrm{S}(1,0,-1) \\ & \overrightarrow{\mathrm{PQ}}=-\hat{\mathrm{i}}+\hat{\mathrm{j}}+3 \hat{\mathrm{k}}, \overrightarrow{\mathrm{PS}}=-\hat{\mathrm{j}}-2 \hat{\mathrm{k}} \end{aligned} \]

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

\[ =|\hat{\imath}-2 \hat{\jmath}+\hat{k}|=\sqrt{6} \text { Answer(6) } \]

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