The image of the point (3,–1,11) in the linex2=y−23=z−34 is

Any point on the given line is(2t,2+3t,3+4t) .  It is the foot of perpendicular  from (3,–1,11)if (2t−3)×2+(2+3t+1)×3+(3+4t−11)×4=0⇒t=1 So, the coordinates of foot of perpendicular are (2,5,7)Now, the point (a, b, c) will be the image of the point (3,–1,11) in the given line if 3+a2=2,b−12=5;c+112=7⇒a=1,b=11,c=3