Answer :

we know that equation of plane passing through (x_{1}, y_{1}, z_{1}) is given by

a(x – x_{1}) + b(y – y_{1}) + c(z – z_{1}) = 0 …… (1)

since, required plane contain lines

and

So, required plane passes through (4, 3, 2) and (3, – 2, 0) so equation of required plane is

a(x – 4) + b(y – 3) + c(z – 2) = 0 …… (2)

plane (2) also passes through (3, – 2, 0), so

a(3 – 4) + b(– 2 – 3) + c(0 – 2)

– a – 5b – 2c = 0

a + 5b + 2c = 0 …… (3)

now plane (2) is also parallel to line with direction ratios 1, – 4, 5 so,

a_{1}a_{2} + b_{1}b_{2} + c_{1}c_{2} = 0

(a)(1) + (b)(– 4) + (c)(5) = 0

a – 4b + 5c = 0 …… (4)

solving equation (3) and (4) by cross – multiplication,

Multiplying by 3,

a = 11 λ, b = – λ, c = – 3 λ

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

a(x – 4) + b(y – 3) + c(z – 2) = 0

(11 λ)(x – 4) + (– λ)(y – 3) + (– 3 λ)(z – 2) = 0

11 λx – 44 λ – λy + 3 λ – 3 λz + 6 λ = 0

11 λx – λy – 3 λz – 35 λ = 0

Dividing by λ,

11x – y – 3z – 35 = 0

So, equation of required plane is 11x – y – 3z – 35 = 0

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