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 (0,0,0), so the equation of plane is

a(x – 0) + b(y – 0) + c(z – 0) = 0

⇒ ax + by + cz = 0 …… (1)

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

x + 2y – z = 1 and 3x – 4y + z = 5

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,

a + 2b – c = 0 …… (b)

3a – 4b + c = 0 …… (c)

Solving (b) and (c) we get,

∴a = – 2λ, b = – 4λ, c = – 10λ

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

(– 2λ)x + (– 4λ)y + (– 10λ)z = 0

Dividing both sides by (– 2λ) we get

x + 2y + 5z = 0

So, the equation of the required planes is x + 2y + 5z = 0