Q50 / 75 · JEE Main 2026 (24 January Shift 1) · Mathematics
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 $y z-$ plane at the point Q. Another line parallel to L and passing through the point $\mathrm{S}(1,0,-1)$ intersects the $y z$-plane at the point R. Then the square of the area of the parallelogram PQRS is equal to $\_\_\_\_$.
Correct answer: 6
Solution
Line L passes through P(1,1,1) perpendicular to two lines with directions $(4, 1, 1)$ and $(1, 1, 0)$.
Direction of L: $(4, 1, 1) \times (1, 1, 0) = (-1, 1, 3)$
Line L: $(x, y, z) = (1, 1, 1) + t(-1, 1, 3)$
Q is intersection with yz-plane (x=0): Setting $1-t=0$ gives $t=1$, so Q = (0, 2, 4)
Line through S(1,0,-1) parallel to L: $(x, y, z) = (1, 0, -1) + s(-1, 1, 3)$
R is intersection with yz-plane: Setting $1-s=0$ gives $s=1$, so R = (0, 1, 2)
For parallelogram PQRS: $\vec{PQ} = (-1, 1, 3)$ and $\vec{PS} = (0, -1, -2)$
Area = $|\vec{PQ} \times \vec{PS}| = |(1, -2, 1)| = \sqrt{1+4+1} = \sqrt{6}$
Square of area = 6
Direction of L: $(4, 1, 1) \times (1, 1, 0) = (-1, 1, 3)$
Line L: $(x, y, z) = (1, 1, 1) + t(-1, 1, 3)$
Q is intersection with yz-plane (x=0): Setting $1-t=0$ gives $t=1$, so Q = (0, 2, 4)
Line through S(1,0,-1) parallel to L: $(x, y, z) = (1, 0, -1) + s(-1, 1, 3)$
R is intersection with yz-plane: Setting $1-s=0$ gives $s=1$, so R = (0, 1, 2)
For parallelogram PQRS: $\vec{PQ} = (-1, 1, 3)$ and $\vec{PS} = (0, -1, -2)$
Area = $|\vec{PQ} \times \vec{PS}| = |(1, -2, 1)| = \sqrt{1+4+1} = \sqrt{6}$
Square of area = 6
Hence, the correct answer is 6.