Answer :

The equation of a plane passing through (x1,y1,z1) is given by

A(x - x1) + B(y - y1) + C(z - z1) = 0


where, A, B, C are the direction ratios of normal to the plane.


Now the plane passes through (-1,3,2)


So, equation of plane is


A(x + 1) + B(y - 3) + C(z - 2) = 0 …(1)


Since this plane is perpendicular to the given two planes.


So, their normal to the plane would be perpendicular to normals of both planes.


we know that -


is perpendicular to both &


So, required normal is cross product of normals of planes


x + 2y + 3z = 5 and 3x + 3y + z = 0


Required Normal





Hence, direction ratios = -7, 8, -3


A = -7, B = 8, C = -3


Putting above values in (1), we get -


A(x + 1) + B(y - 3) + C(z - 2) = 0


-7(x + 1) + 8(y - 3) + (-3)(z - 2) = 0


-7x - 7 + 8y - 24 - 3z + 6 = 0


-7x + 8y - 3z - 25 = 0


7x - 8y + 3z + 25 = 0


Therefore, equation of the required plane is 7x - 8y + 3z + 25 = 0.


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