Find the equation of the plane through the line of intersection of the planes x + 2y + 3z + 4 = 0 and x – y + z + 3 = 0 and passing through the origin ?
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
So equation of plane passing through the line of intersection of given two planes
x + 2y + 3z + 4 = 0 and x – y + z + 3 = 0 is
(x + 2y + 3z + 4) + k(x – y + z + 3) = 0
x(1 + k) + y(2 – k) + z(3 + k) + 4 + 3k = 0 …… (1)
Equation (1) is passing through origin , so
(0)(1 + k) + (0)(2 – k) + (0)(3 + k) + 4 + 3k = 0
0 + 0 + 0 + 4 + 3k = 0
3k = – 4
Put the value of k in equation (1),
x(1 + k) + y(2 – k) + z(3 + k) + 4 + 3k = 0
x(1
) + y(2 + 
) + z(3
) + 4
= 0
x(
) + y(
) + z(
) + 4
= 0
Multiplying by 3, we get
– x + 10y + 5z = 0
x – 10y – 5z = 0
the equation of required plane is, x – 10y – 5z = 0