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Question
Find the equation of the plane determined by the intersection of the lines \[\frac{x + 3}{3} = \frac{y}{- 2} = \frac{z - 7}{6} \text{ and }\frac{x + 6}{1} = \frac{y + 5}{- 3} = \frac{z - 1}{2}\]
Solution
\[\text{ The given equations of the lines are} \]
\[\frac{x + 3}{3} = \frac{y}{- 2} = \frac{z - 7}{6} . . . \left( 1 \right)\]
\[\frac{x + 6}{1} = \frac{y + 5}{- 3} = \frac{z - 1}{2} . . . \left( 2 \right)\]
\[\text{ Let the direction ratios of the plane be proportional to a, b, c . } \]
\[\text{ Since the plane contains line (1), it should pass through (-3, 0, 7) and is parallel to the line (1) } .\]
\[\text{ Equation of the plane through (1) is} \]
\[a \left( x + 3 \right) + b \left( y \right) + c \left( z - 7 \right) = 0 . . . \left( 3 \right), \]
\[\text{ where } 3a - 2b + 6c = 0 . . . \left( 4 \right)\]
\[\text{ Since the plane contains line (2), the plane is parallel to line (2) also. } \]
\[ \Rightarrow a - 3b + 2c = 0 . . . \left( 5 \right)\]
\[\text{ Solving (4) and (5) using cross-multiplication, we get } \]
\[\frac{a}{14} = \frac{b}{0} = \frac{c}{- 7}\]
\[\text{ Substitutinga, b and c in (3), we get} \]
\[14 \left( x + 3 \right) + 0 \left( y \right) - 7 \left( z - 7 \right) = 0\]
\[ \Rightarrow 2 \left( x + 3 \right) + 0 \left( y \right) - 1 \left( z - 7 \right) = 0\]
\[ \Rightarrow 2x - z + 13 = 0\]
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