# 13. Find the equation of the plane passing through the point (– 1, 3, 2) and perpendicular to each of the planes x + 2y + 3z = 5 and 3x + 3y + z = 0.

Given

two planes x + 2y + 3z = 5 and 3x + 3y + z = 0.

the normal vectors  of these plane are

Since the normal vector of the required plane is perpendicular to the normal vector of given planes, the required plane's normal vector will be :

Now, as we know

the equation of a plane in vector form is :

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

Hence the equation of the plane is

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