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Question

Find the equation of the plane that contains the point (1,1,2) and is perpendicular to each of the planes 2x+3y2z=5 and x+2y3z=8.

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Solution

Solution The equation of the plane containing the given point is
A(x1)+B(y+1)+C(z2)=0...(1)
Applying the condition of perpendicularity to the plane given in (1) with the planes
2x+3y2z=5 and x+2y3z=8, we have
2A+3B2C=0 and A+2B3C=0
Solving these equations, we find A=5C and B=4C. Hence, the required equation is
5C(x1)+4C(y+1)+C(z2)=0
i.e 5x4yz=7

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