we know that equation of plane passing through (x₁, y₁, z₁) is given by

a(x – x₁) + b(y – y₁) + c(z – z₁) = 0 …… (1)

Required plane is passing through (0, 7, – 7) so

a(x – 0) + b(y – 7) + c(z + 7) = 0

ax + b(y – 7) + c(z + 7) = 0 …… (2)

plane (2) also contain line so, it passes through point (– 1, 3, – 2),

a(– 1) + b(3 – 7) + c(– 2 + 7) = 0

– a – 4b + 5c = 0 …… (3)

Also plane (2) will be parallel to line so,

a₁a₂ + b₁b₂ + c₁c₂ = 0

(a)(– 3) + (b)(2) + (c)(1) = 0

– 3a + 2b + c = 0 …… (4)

Solution (3) and (4) by cross – multiplication,

a = – 14, b = – 14, c = – 14

put a, b, c in equation (2),

ax + b(y – 7) + c(z + 7) = 0

(– 14)x + (– 14)(y – 7) + (– 14)(z + 7) = 0

Dividing by (– 14) we get

x + y – 7 + z + 7 = 0

x + y + z = 0

so, equation of plane containing the given point and line is x + y + z = 0

the other line is

so, a₁a₂ + b₁b₂ + c₁c₂ = 0

(1)(1) + (1)(– 3) + (1)(2) = 0

1 – 3 + 2 = 0

0 = 0

LHS = RHS

So, lie on plane x + y + z = 0