CGPET 2022 Solved Paper

Let
L1:x+3∕−3=y−3∕2=z−6∕4=r1
and
L2:x−5∕3=y−7∕−1=z+2∕1=r2

Again, let
A and B are the points of intersection of the line with L1 and L2. Then,
A(−3r1−3,2r1+3,4r1+6) and
B(3r2+5,−r2+7,r2−2)
Now, DR′s of AB are
<3r2+5+3r1+3,−r2+7−2r1−3,r2−2−4r1−6>
i.e., <3r2+3r1+8,−r2−2r1+4,r2−4r1−8>
But DR′s o f AB are <2,7,−5>.
∴

3r2+3r1+8

2

=

−r2−2r1+4

7

=

r2−4r1−8

−5

⇒7(3r2+3r1+8)=2(−r2−2r1+4)
and −5(−r2−2r1+4)=7(r2−4r1−8)


‌⇒23r2+25r1=−48
and −2r2+38r1=−36
∴r1=−1and r2=−1

So, coordinates of A are
(0,1,2)
and B are
(2,8,−3)
∴ Length of intercept,
AB=√(2−0)2+(8−1)2+(−3−2)2
=√4+49+25=√78