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

The given plane is x−y−z=9-----(1) The given line AB is x−22=y+23=z−6−6-------(2) The equation of the line passing through (1,0,−3) and parallel  to x−22=y+23=z−6−6 is x−12=y−03=z+3−6=r-----(3) Coordinate of any point on (3) may be given as P(2r+1 , 3r,−6r−3) If P is the point of the intersection of 1 and,3 then it must  lie on (1). Therefore, (2r+1)−(3r)−(−6r−3)=92r+1−3r+6r+3=9 or r=1 Therefore, the coordinates of P are 3,3,−9   Distance between Q(1,0,−3) and P(3,3,−9)=(3−1)2+(3−0)2+(−9+3)2=4+9+36=7