Find the distance of the point (1, 0, -3) from the plane x - y -z = 9 measured parallel to the line $\frac{\mathrm{x}-2}{2}=\frac{\mathrm{y}+2}{3}=\frac{\mathrm{z}-6}{-6}$ 

Given plane is x - y - z = 9---------(1)

Given line AB is $\frac{\mathrm{x}-2}{2}=\frac{\mathrm{y}+2}{3}=\frac{\mathrm{z}-6}{-6}---\left(2\right)$

Equation of a line passing through the point Q (1, 0, -3) and parallel to line (2) is $\frac{\mathrm{x}-1}{2}=\frac{\mathrm{y}}{3}=\frac{\mathrm{z}+3}{-6}=\mathrm{r}-----\left(3\right)$

Co-ordinates of point on line (3) may be taken as P (2r + 1, 3r, -6r -3)
If P is the point of intersection of line (3) and plane (1), then P lies on plane (1),

(2r +1)- (3r) - (- 6r-3) = 9 gives r = 1

or  $\mathrm{P}\equiv (3,3,-9)$

Distance between points Q(1, 0, -3) and P (3, 3, - 9)

$\mathrm{PQ}=\sqrt{(3-1{)}^2+(3-0{)}^2+(-9-(-3){)}^2}=\sqrt{4+9+36}=7$