Q. 12

# Find the equation

We know that solution of a plane passing through (x1,y1,z1) is given as -

a(x – x1) + b(y – y1) + c(z – z1) = 0

The required plane passes through (1, – 1,2), so the equation of plane is

a(x – 1) + b(y + 1) + c(z – 2) = 0

ax + by + cz = a – b + 2c …… (1)

Now, the required plane is also perpendicular to the planes,

2x + 3y – 2z = 5 and x + 2y – 3z = 8

We know that planes a1x + b1y + c1z + d1 = 0 and a2x + b2y + c2z + d2 = 0 are at right angles if,

a1a2 + b1b2 + c1c2 = 0 …… (a)

Using (a) we have,

2a + 3b – 2c = 0 …… (b)

a + 2b – 3c = 0 …… (c)

Solving (b) and (c) we get,

a = – 5λ, b = 4λ, c = λ

Putting values of a,b,c in equation (1) we get,

(– )x + (4λ)y + (λ)z = – 5λ – 4λ + 2λ

– 5λx + 4λy + λz = – 7λ

Dividing both sides by (– λ) we get

5x – 4y – z = 7

So, the equation of the required planes is 5x - 4y – z = 7

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