Match the following listsa.  The distance between the line  r→=(2i^−2j^+3k^)+λ(i^−j^+4k^)  and      plane r→⋅(i^+5j^+k^)=5d. The distance of the point (1,0, - 3) from the plane x-y -z - 9=0 measured parallel     to x−22=y+23=z−6−6

a. The given line and plane are  r→=(2i^−2j^+3k^)+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 is equal to 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 distance between two parallel planes (2, 5,- 3) from the plane r→⋅(6i−3j+2k)=4 or 6x-3y+22 -4=0 is      d=|12−15−6−4|36+9+4=13/49=13/7d.  The equation of the line AB is x−12=y3=z+3−6=r            The coordinates of any point on line P(2r+1, 3r, - 6r - 3) which lie on plane       (2r+1)−(3r)−(−6r−3)=9r=1point  P≡(3,3,−9)      Required distance      PQ=(3−1)2+(3−0)2+(−9+3)2=4+9+36=7