Find the distance of the point (1, 0, -3) from the plane x - y -z = 9 measured parallel to the line x−22=y+23=z−6−6

Given plane is x - y - z = 9---------(1)Given line AB is x−22=y+23=z−6−6---2Equation of a line passing through the point Q (1, 0, -3) and parallel to line (2) is x−12=y3=z+3−6=r-----3Co-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 = 1or  P≡(3,3,−9)Distance between points Q(1, 0, -3) and P (3, 3, - 9)PQ=(3−1)2+(3−0)2+(−9−(-3))2=4+9+36=7