We know that equation of plane passing through the intersection of planes a₁x + b₁y + c₁z + d₁ = 0 and a₂x + b₂y + c₂z + d₂ = 0 is given by

(a₁x + b₁y + c₁z + d₁) + k(a₂x + b₂y + c₂z + d₂) = 0

So, equation of plane passing through the intersection of planes

3x – 4y + 5z – 10 = 0 and 2x + 2y – 3z – 4 = 0 is

(3x – 4y + 5z – 10) + k(2x + 2y – 3z – 4) = 0 ……(1)

⇒ x(3 + 2k) + y(– 4 + 2k) + z(5 – 3k) + (– 10 – 4k) = 0

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

Given the plane is parallel to line

6×(3 + 2k) + 3× (– 4 + 2k) + 2×(5 – 3k) = 0

⇒ 18 + 12k – 12 + 6k + 10 – 6k = 0

⇒ k =

Putting the value of k in equation (1)

⇒

⇒

⇒ x – 20y + 27z – 14 = 0

The required equation is x – 20y + 27z – 14 = 0