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Equation of a Plane Parallel to a Given Plane

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Question

Find the equation of a plane which passes through the point $(1, 2, 3)$ and which is equally inclined to the planes $x - 2 y + 2 z - 3 = 0$ and $8 x - 4 y + z - 7 = 0$ .

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Solution

Equation of angle bisector of a given plane:
$\begin{matrix} \frac{x - 2 y + 2 z - 3}{\sqrt{9}} = \pm \frac{(8 x - 4 y + z - 7)}{\sqrt{64 + 16 + 7}} \frac{x - 2 y + 2 z - 3}{3} = \pm \frac{(8 x - 4 y + z - 7)}{9} 3 (x - 2 y + 2 z - 3) = 8 x - 4 y + z - 7 5 x - 5 z + 2 = 0 t a k i n g - v e s i g n 3 (x - 2 y + 2 z - 3) = - (8 x - 4 y + z - 7) 11 x - 10 y + 7 z - 16 = 0 E q u a t i o n o f a p l a n e i s : 5 x - 5 z + c_{1} = 0 5 - 5 (3) + c_{1} = 0 c_{1} = 10 5 x - 5 z + 10 = 0 11 x - 10 y + 7 z + c_{2} = 0 11 - 20 + 21 + c_{2} = 0 c_{2} = - 12 11 x - 10 y + 7 z - 12 = 0 \end{matrix}$

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Q. Find the equation of the sphere which passes through the circle

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