The equations of line BC are (using two point form)
x−1
5−1
=
y−4
4−4
=
z−6
4−6
i.e.
x−1
2
=
y−4
0
=
z−6
−1
Any point on this line is (2t+1,4,−t+6) . If this is the foot of perpendicular from A onthe line BC, then d.n. of this perpendicular are <2t+1−1,4−1,−t+6−1> i.e. <2t,3,−t+5> Using condition of perpendicularity, we have (2t)(2)+3×0+(−t+5)(−1)=0 ⇒5t−5=0⇒t=1 ∴ Required foot of perpendicular is (2+1,4,−1+6) =(3,4,5)