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Question
Find the equation of the plane through the points (2, 2, −1) and (3, 4, 2) and parallel to the line whose direction ratios are 7, 0, 6.
Solution
\[\text{ The equation of the plane through (2, 2, -1) is}\]
\[a \left( x - 2 \right) + b \left( y - 2 \right) + c \left( z + 1 \right) = 0 . . . \left( 1 \right)\]
\[\text{ This plane passes through (3, 4, 2). So } ,\]
\[a \left( 3 - 2 \right) + b \left( 4 - 2 \right) + c \left( 2 + 1 \right) = 0\]
\[ \Rightarrow a + 2b + 3c = 0 . . . \left( 2 \right) \]
\[\text{ Again plane (1) is parallel to the line whose direction ratios are 7, 0, 6 } .\]
\[\text{ It means that the normal of plane (1) is perpendicular to theline whose direction ratios are 7, 0, 6 } . \]
\[ \Rightarrow 7a + 0b + 6c = 0 (\text{ Because } a_1 a_2 + b_1 b_2 + c_1 c_2 = 0)\]
\[\text{ Solving (1), (2) and (3), we get} \]
\[\begin{vmatrix}x - 2 & y - 2 & z + 1 \\ 1 & 2 & 3 \\ 7 & 0 & 6\end{vmatrix} = 0\]
\[ \Rightarrow 12 \left( x - 2 \right) + 15 \left( y - 2 \right) - 14 \left( z + 1 \right) = 0\]
\[ \Rightarrow 12x + 15y - 14z -68 = 0\]
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