Question

find the distance of the point (1,-2,3) from the plane x-y+z=5 measured parallel to the line x/2=y/3=z/-6

Answer 1

start by writing the equation of line parallel to the given line,
and passing thru the given point

Answer 2

parametric form of line passing thru (1, -2, 3) and parallel to the given line :
\(\large \mathbb{x = 2t+1}\)
\(\large \mathbb{y = 3t-2}\)
\(\large \mathbb{z = -6t+3}\)

Answer 3

next, find at what point this line intersects the given plane

Answer 4

lastly find the distance using the distance formula

Answer 5

I'll like to modify ganeshie's solution a li'll bit.

As we are working with parametric form; we only need to use the condition that the point
(2t+1, 3t-2, -6t+3) satisfies the equation of plane x-y+z=5

The absolute value of 't' will be the required distance itself.

PS: I am lazy ;)