Given: A = (1, 0, 3) B = (4, 7, 1) C = (3, 5, 3) Direction ratio of line BC is (-1, -2, 2) Equation of BC,
x−4
3−4
=
y−7
5−7
=
z−1
3−1
⇒
x−4
−1
=
y−7
−2
=
z−1
2
=k ⇒x=−k+4,y=−2k+7 and z=2k+1 General point on BC be D = (-k + 4, -2k + 7, 2k + 1) and let D be foot of perpendicular. Direction ratio of line AD will be (-k + 4 - 1), (-2k + 7 - 0), (2k + 1 - 3) ⇒ (-k + 3, -2k + 7, 2k - 2) Now, since the line BC and AD are perpendicular, l1l2+m1m2+nln2=0 So, (−k+3)×−1+(−2k+7)×−2+(2k−2)×2=0 ⇒k−3+4k−14+4k−4=0 ⇒9k=21 ⇒k=