# Find the equation

We know that equation of plane passing through the intersection of 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 intersection of planes

3x – 4y + 5z – 10 = 0 and 2x + 2y – 3z – 4 = 0 is

(3x – 4y + 5z – 10) + k(2x + 2y – 3z – 4) = 0 ……(1)

x(3 + 2k) + y(– 4 + 2k) + z(5 – 3k) + (– 10 – 4k) = 0

We know that line is parallel to plane a2x + b2y + c2z + d2 = 0 if a1a2 + b1b2 + c1c2 = 0

Given the plane is parallel to line 6×(3 + 2k) + 3× (– 4 + 2k) + 2×(5 – 3k) = 0

18 + 12k – 12 + 6k + 10 – 6k = 0

k = Putting the value of k in equation (1)   x – 20y + 27z – 14 = 0

The required equation is x – 20y + 27z – 14 = 0

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