KCET 2015 Math Solved Paper

Given equation of planes,
P1:x−2y−z+5=0 → (1)
P2:x+y+3z=6 → (2)
and point P(2, 3, 1).
Normal vector of Eq. (1) is given by,

→

N

1=

∧

i

−2

∧

j

+

∧

k

→ (3)
Similarly, normal vector of Eq. (2) is given by

→

N

2=

∧

i

+

∧

j

+3

∧

k

→ (4)
Vector perpendicular to the normal to the plans are:
So, required equation of line passing through the point (2,3,1) is,
‌

x−2

−5

=‌

y−3

−4

=‌

z−1

3