# Find the equation

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, Rate this question :

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