Question

Find the equation of the plane passing through the points (2, 2, 1) and (9, 3, 6) and perpendicular to the plane 2x + 6y + 6z = 1.

 

Answer 1

\[\text{ The equation of any plane passing through (2, 2, 1) is } \]
\[a \left( x - 2 \right) + b \left( y - 2 \right) + c \left( z - 1 \right) = 0 . . . \left( 1 \right)\]
\[ \text{ It is given that (1) is passing through (9, 3, 6). So,} \]
\[a \left( 9 - 2 \right) + b \left( 3 - 2 \right) + c \left( 6 - 1 \right) = 0\]
\[ \Rightarrow 7a + b + 5c = 0 . . . \left( 2 \right)\]
\[ \text{ It is given that (1) is perpendicular to the plane 2x + 6y + 6z = 1 . So,}\]
\[2a + 6b + 6c = 0\]
\[ \Rightarrow a + 3b + 3c = 0 . . . \left( 3 \right)\]
\[ \text{ Solving (1), (2) and (3), we get } \]
\[\begin{vmatrix}x - 2 & y - 2 & z - 1 \\ 7 & 1 & 5 \\ 1 & 3 & 3\end{vmatrix} = 0\]
\[ \Rightarrow - 12 \left( x - 2 \right) - 16 \left( y - 2 \right) + 20 \left( z - 1 \right) = 0\]
\[ \Rightarrow 3 \left( x - 2 \right) + 4 \left( y - 2 \right) - 5 \left( z - 1 \right) = 0\]
\[ \Rightarrow 3x + 4y - 5z = 9\]