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

a).The given line and plane are r→=(2i^−2j^+2k^)+λ(i^−j^+4k^) and r→⋅(i^+5j^+k^)=5 respectively. Since (i^−j^+4k^)⋅(i^+5j^+k^)=0  line and plane are parallel.Hence, the required distance = distance of point (2,-2, 3) from the plane x+ 5y+z-5 -0       Which is |2−10+3−5|1+25+1=1033b). The distance between two parallel planes r→⋅(2i−j+3k)=4 and r→⋅(6i−3j+9k)+13=0 is      d=|4−(−13/3)|(2)2+(−1)2+(3)2=(25/3)14=25314c) The perpendicular distance of the point (2, 5,-3) from the plane r→⋅(6i−3j+2k)=4 or 6x−3y+2z−4=0 is     d=|12−15−6−4|36+9+4=1349=137d) The equation of the line AB is x−22=y+23=z−6−6 The equation of line passing through (1, 0, -3) and parallel to AB is     x−12=y3=z+3−6=r (say)            The coordinates of any point on line P(2r+1,3r,-6r-3) which plane(2r+1) - 3r - ( -6r - 3)=9  ⇒  r =1      Required distance       PQ=(3−1)2+(3−0)2+(−9+3)3=4+9+36=7