VITEEE 2018 Paper

Since The line

x−2

3

=

y−1

−5

=

z+2

2

lie in the plane
x + 3y - αz + β = 0
∴ Pt (2 , - 1 , - 2) lies on the plane i.e. 2 + 3 + 2α + β = 0 or 2α + β + 5 = 0 ... (i)
Also normal to plane will be perpendicular to line,
∴ 3 × 1 - 5 × 3 + 2 × (-α) = 0 ⇒ α = - 6
From equation (i) then, β = 7
∴ (α , β) = (- 6 , 7)