Find the equation of the plane through the points (2, 2, – 1) and (3, 4, 2) and parallel to the line whose direction ratios are 7, 0, 6.

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

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

So, equation of plane passing through (2,2, – 1) is

a(x – 2) + b(y – 2) + c(z + 1) = 0 ……(2)

It also passes through (3,4,2)

So, equation (2) must satisfy the point (3,4,2)

∴ a(3 – 2) + b(4 – 2) + c(2 + 1) = 0

⇒ a + 2b + 3c = 0 ……(3)

We know that line is parallel to plane a₂x + b₂y + c₂z + d₂ = 0 if a₁a₂ + b₁b₂ + c₁c₂ = 0 ……(4)

Here, the plane (2) is parallel to line having direction ratios 7,0,6 ,

So,

a×7 + b×0 + c×6 = 0

⇒ 7a + 6c = 0

⇒ ……(5)

Putting the value of a in equation (3)

a + 2b + 3c = 0

⇒ + 2b + 3c = 0

⇒ – 6c + 14b + 21c = 0

⇒ 14b + 15c = 0

⇒

Putting the value of a and b in equation (2)

a(x – 2) + b(y – 2) + c(z + 1) = 0

⇒ (x – 2) + (y – 2) + c(z + 1) = 0

⇒

Multiplying by we have,

– 12x + 24 – 15y + 30 + 14z + 14 = 0

⇒ – 12x + 15y + 14z + 68 = 0

⇒ 12x – 15y – 14z – 68 = 0

Equation of required plane is 12x – 15y – 14z – 68 = 0