The distance of the point (1,-5,9) from the plane x-y+z=5 measured along a straight line x=y=z is

The distance of the point (1,-5,9) from the plane $x - y + z = 5$ measured along a straight line $x = y = z$ is

A
$3 \sqrt{5}$
B
$10 \sqrt{3}$
C
$5 \sqrt{3}$
D
$3 \sqrt{10}$

Solution:

Let $P = (1, - 5, 9)$

Let Q be a point on the given plane such that

PQ is parallel to given line

The equation of the line PQ is $\frac{x - 1}{1} = \frac{y + 5}{1} = \frac{z - 9}{1}$

Let $Q = (1 + t, t - 5, t + 9)$

Sub Q in the given plane, $t = - 10$

$∴ Q (- 9, - 15, - 1) P Q = \sqrt{300} = 10 \sqrt{3}$

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