MHT CET 2020 Math Solved Paper

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)