The distance of the plane 3x+4y+5z+19=0 from the point (1,-1,1) measured along a line parallel to the line with direction ratios 2,3,1 is
2352
7152
14
23
Line passing through (1,-1,1) is x−12=y+13=z−11=t
P(2t+1,3t−1,t+1) lies on 3x+4y+5z+19=0
t=−1⇒P(−1,−4,0)Q(1,−1,1)
PQ=14