Let N be the foot of the perpendicular from the point P(0,2,3) on the given line ‌
x+3
5
=‌
y−1
2
=‌
z+4
3
=r (say) . . . (i) Any point on the line (i) is (5r−3,2r+1,3r−4)‌. ‌ If this point is N, The direction ratio's of NP are i.e. ‌‌5r−3,2r−1,3r−7 Since, NP is perpendicular to the given line, then ‌‌‌5(5r−3)+2(2r−1)+3(3r−7)=0 ‌⇒‌‌38r−38=0⇒r=1 ‌‌ The point ‌N‌ is ‌‌‌(5−3,2+1,3−4) ‌‌ i.e., ‌‌‌(2,3,−1).