AP EAPCET 04-Jul-2022 Shift 2 Paper

(b) Given, Q(2,2,1) and R(5,2,−2)
Thus, equation of line through R and Q is
‌

x−2

2−5

‌=‌

y−2

2−2

=‌

z−1

1+2

⇒‌

x−2

−3

=‌

y−2

0

=‌

z−1

3

=r‌ (say) ‌
Thus, P be point on the line. So,
P=(−3r+2),2,(3r+1)
Since, x - coordinate is 4 . So, −3r+2=4
⇒‌‌r=‌

−2

3

Thus, y-coordinate =2
and z-coordinate =−1
∴y - coordinate =−2(z− coordinate of P).