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 (0,0,0) is

a(x – 0) + b(y – 0) + c(z – 0) = 0

ax + by + cz = 0 ……(2)

It also passes through (3, – 1,2)

So, equation (2) must satisfy the point (3, – 1,2)

∴ 3a – b + 2c = 0 ……(3)

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

Here, the plane is parallel to line,

So,

a×1 + b× – 4 + c×7 = 0

⇒ a – 4b + 7c = 0 ……(5)

Solving equation (3) and (5) by cross multiplication we have,

⇒

⇒

∴ a = k, b = – – 19k and c = – 11k

Putting the value in equation (2)

ax + by + cz = 0

kx – 19ky – 11kz = 0

Dividing by k we have

x – 19y – 11z = 0

The required equation is x – 19y – 11z = 0