If (a, b, c) is the image of the point (1, 2, -3) in the line x+12=y−3−2=z−1then  a + b + +c is equal to

So, foot of perpendicular will be a+12,b+22,c−32Let foot of perpendicular on x+12=y−3−2=z−1beF2t−1,3−2t,−tfrom  A1,2,−3AFwill be perpendicular to line with direction ratios <2,−2,−1>which gives the relation  22t−1−1−23−2t−2−1−t+3=0⇒9t−9=0⇒t=1So, foot of perpendicular will be F1,1,−1which should be same as a+12,b+22,c−32Therefore, ⇒a=1,b=0,c=1