If a plane passes through a straight line "3x + 4y-2z + 5 = 0; x-2y + Z + 7 = 0;", and the intercept on the z-axis is - 3, its equation is solved

If a plane passes through a straight line "3x + 4y-2z + 5 = 0; x-2y + Z + 7 = 0;", and the intercept on the z-axis is - 3, its equation is solved

Plane beam equation method
Let the plane equation passing through the line be:
(3x+4y-2z+5)+k(x-2y+z+7)=0
Substituting x = 0, y = 0, z = - 3, we can get k = - 11 / 4
So the plane equation is as follows:
x+38y-19z-57=0.
Give some encouragement, thank you!