The distance of the point (1, –2, 3) from the plane x – y + z = 5 measured parallel to the line $\frac{x}{2}=\frac{y}{3}=\frac{z}{-6}$ is :

1 unit

Here we are not to find perpendicular distance of the point from the plane but distance measured along with the given line. The method is as follow:

The equation of the line through the point (1, –2, 3) and parallel to given line is 

$\frac{x-1}{2}=\frac{y+2}{3}=\frac{z-3}{-6}$ = r (say)

The coordinate of any point on it is (2r + 1, 3r – 2, –6r + 3).

If this point lies in the given plane then

2r + 1 – (3r – 2) + (–6r + 3) = 5 ⇒ –7r = –1 or r = $\frac{1}{7}$

∴ point of intersection is $\left(\frac{9}{7},-\frac{11}{7}, \frac{15}{7}\right)$

∴ The required distance

= the distance between the points (1, –2, 3) and $\left(\frac{9}{7},-\frac{11}{7}, \frac{15}{7}\right)$

$=\sqrt{\left(1-\frac{9}{7}\right)^2+\left(-2+\frac{11}{7}\right)^2+\left(3-\frac{15}{7}\right)^2}=\frac{1}{7} \sqrt{49}$ = 1 unit