# Find the equation

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

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