Find the equation of the plane through the intersection of the planes 2x – 7y + 4z – 3 = 0 and 3x – 5y + 4z + 11 = 0 and the point ( – 2,1,3)?
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