we know that, equation of a plane passing through the line of intersection of two 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

Given , equation of plane is,

2x – 7y + 4z – 3 = 0 and 3x – 5y + 4z + 11 = 0

So equation of plane passing through the line of intersection of given two planes is

(2x – 7y + 4z – 3) + k(3x – 5y + 4z + 11) = 0

2x – 7y + 4z – 3 + 3kx – 5ky + 4kz + 11k = 0

x(2 + 3k) + y( – 7 – 5k) + z(4 + 4k) – 3 + 11k = 0 ……(1)

As given that, plane (1) is passing through the point ( – 2,1,3) so it satisfy the equation (1),

( – 2)(2 + 3k) + (1)( – 7 – 5k) + (3)(4 + 4k) – 3 + 11k = 0

– 2 + 12k = 0

12k = 2

put the value of k in equation (1)

x(2 + 3k) + y( – 7 – 5k) + z(4 + 4k) – 3 + 11k = 0

x(2 + ) + y( – 7 – ) + z(4 + ) – 3 + = 0

x() + y() + z() = 0

x() + y() + z() = 0

multiplying by 6 , we get

15x – 47y + 28z – 7 = 0

Therefore , equation of required plane is 15x – 47y + 28z – 7 = 0