Question

Find the equation of the plane passing through the points (3, 4, 1) and (0, 1, 0) and parallel to the line 

\[\frac{x + 3}{2} = \frac{y - 3}{7} = \frac{z - 2}{5} .\]

  

Answer 1

\[ \text{ The equation of the plane passing through the point (3,4, 1) is } \]
\[a \left( x - 3 \right) + b \left( y - 4 \right) + c \left( z - 1 \right) = 0 . . . \left( 1 \right)\]
\[ \text{ This plane passes through (0, 1, 0). So } ,\]
\[a \left( 0 - 3 \right) + b \left( 1 - 4 \right) + c \left( 0 - 1 \right) = 0\]
\[ \Rightarrow - 3a - 3b - c = 0\]
\[ \Rightarrow 3a + 3b + c = 0 . . . \left( 2 \right)\]
\[ \text{ Again plane (1) is parallel to the given line } .\]
\[ \text{ It means that the normal to plane (1) is perpendicular to the line. } \]
\[ \Rightarrow a \left( 2 \right) + b \left( 7 \right) + c \left( 5 \right) = 0 . . . \left( 3 \right) (\text{ Because } a_1 a_2 + b_1 b_2 + c_1 c_2 = 0)\]
\[\text{ Solving (1), (2) and (3), we get } \]
\[\begin{vmatrix}x - 3 & y - 4 & z - 1 \\ 3 & 3 & 1 \\ 2 & 7 & 5\end{vmatrix} = 0\]
\[ \Rightarrow 8 \left( x - 3 \right) - 13 \left( y - 4 \right) + 15 \left( z - 1 \right) = 0\]
\[ \Rightarrow 8x - 13y + 15z + 13 = 0\]