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JEE Main · 2026 · 24 Jan 2026 · Shift 1 · Three Dimensional Geometry

Let a line L passing through the point

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

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 ______.

Answer: 6

Solution

The direction of

L

is perpendicular to

(4, 1, 1)

and

(1, 1, 0)

, so

d_{L} = (4, 1, 1) \times (1, 1, 0) = (- 1, 1, 3)

. Through

P (1, 1, 1)

,

L

meets the

y z

-plane (

x = 0

) at parameter

1

, giving

Q (0, 2, 4)

. The parallel line through

S (1, 0, - 1)

meets

x = 0

at

R (0, 1, 2)

. With

P Q = (- 1, 1, 3)

and

P S = (0, - 1, - 2)

,

P Q \times P S = (1, - 2, 1)

, whose length is

6

. The square of the area of parallelogram

P QR S

is

6

.

JEE Main · 2026 · 24 Jan 2026 · Shift 1 · Integer answer

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More JEE Main questions on Three Dimensional Geometry