Find the equation of the plane through the line of intersection of the planes x – 3y + 2z – 5 = 0 and 2x – y + 3z – 1 = 0 and passing through (1, – 2, 3)?
We know that equation of plane passing through the line of intersection of planes
(a1x + b1y + c1z + d1) + k(a2x + b2y + c2z + d2) = 0
So equation of plane passing through the line of intersection of planes
x – 3y + 2z – 5 = 0 and 2x – y + 3z – 1 = 0 is given by
(x – 3y + 2z – 5) + k(2x – y + 3z – 1) = 0
x(1 – 2k) + y( – 3 – k) + z(2 + 3k) – 5 – k = 0 …… (1)
plane (1) is passing the through the point(1, – 2, 3) so,
1(1 + 2k) + ( – 2)( – 3 – k) + (3)(2 + 3k) – 5 – k = 0
1 + 2k + 6 + 2k + 6 + 9k – 5 – k = 0
8 + 12k = 0
12k = – 8
Put value of k in eq.(1),
x(1 + 2k) + y( – 3 – k) + z(2 + 3k) – 5 – k = 0
Multiplying by( – 13),
x + 7y + 13 = 0
Equation of required plane is, 