The distance of the point (1, 0, 2) from the point of intersection of the line (x-2)/3 = (y+1)/4 = (z-2)/12 and the plane x - y + z = 16 is

The distance of the point (1, 0, 2) from the point of… — JEE Main 2015 Mathematics PYQ

JEE Main 2015

The distance of the point

(1, 0, 2)

from the point of intersection of the line

\dfrac{x-2}{3} = \dfrac{y+1}{4} = \dfrac{z-2}{12}

and the plane

x - y + z = 16

A.
$2\sqrt{14}$
B.
$8$
C.
$3\sqrt{21}$
D.
$13$

Show answer and solution

Correct option: D

Solution

Parametrize the line: x = 2+3t, y = -1+4t, z = 2+12t. Substitute into the plane x - y + z = 16: (2+3t) - (-1+4t) + (2+12t) = 16 → 5 + 11t = 16 → t = 1. So the intersection point is (5, 3, 14). Distance from (1, 0, 2): √((5-1)² + (3-0)² + (14-2)²) = √(16 + 9 + 144) = √169 = 13.

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