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Question
The equation of the plane through the intersection of the planes x + 2y + 3z = 4 and 2x + y − z = −5 and perpendicular to the plane 5x + 3y + 6z + 8 = 0 is
Options
7x − 2y + 3z + 81 = 0
23x + 14y − 9z + 48 = 0
51x − 15y − 50z + 173 = 0
None of these
Solution
51x − 15y − 50z + 173 = 0
\[\text{ The equation of the plane passing through the line of intersection of the given planes is } \]
\[x + 2y + 3z - 4 + \lambda \left( 2x + y - z + 5 \right) = 0 \]
\[\left( 1 + 2\lambda \right)x + \left( 2 + \lambda \right)y + \left( 3 - \lambda \right)z - 4 + 5\lambda = 0 . . . \left( 1 \right)\]
\[\text{ This plane is perpendicular to 5x + 3y + 6z + 8 = 0 . So } ,\]
\[5 \left( 1 + 2\lambda \right) + 3\left( 2 + \lambda \right) + 6 \left( 3 - \lambda \right) = 0............... (\text{ Because } a_1 a_2 + b_1 b_2 + c_1 c_2 = 0)\]
\[ \Rightarrow 5 + 10\lambda + 6 + 3\lambda + 18 - 6\lambda = 0\]
\[ \Rightarrow 7\lambda + 29 = 0\]
\[ \Rightarrow \lambda = \frac{- 29}{7}\]
\[\text{ Substituting this in (1), we get } \]
\[\left( 1 + 2 \left( \frac{- 29}{7} \right) \right)x + \left( 2 + \left( \frac{- 29}{7} \right) \right)y + \left( 3 + \frac{29}{7} \right)z - 4 + 5 \left( \frac{- 29}{7} \right) = 0\]
\[ \Rightarrow 51x + 15y - 50z + 173 = 0\]
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