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Question
Find the equation of the plane which contains the line of intersection of the planes x \[+\] 2y \[+\] 3 \[z - \] 4 \[=\] 0 and 2 \[x + y - z\] \[+\] 5 \[=\] 0 and whose x-intercept is twice its z-intercept. Hence, write the equation of the plane passing through the point (2, 3, \[-\] 1) and parallel to the plane obtained above.
Solution
The equation of the family of planes passing through the intersection of the planes x + 2y + 3z − 4 = 0 and 2x + y − z + 5 = 0 is
(x + 2y + 3z − 4) + k(2x + y − z + 5) = 0, where k is some constant
\[\Rightarrow \left( 2k + 1 \right)x + \left( k + 2 \right)y + \left( 3 - k \right)z = 4 - 5k\]
\[ \Rightarrow \frac{x}{\left( \frac{4 - 5k}{2k + 1} \right)} + \frac{y}{\left( \frac{4 - 5k}{k + 2} \right)} + \frac{z}{\left( \frac{4 - 5k}{3 - k} \right)} = 1\]
It is given that x-intercept of the required plane is twice its z-intercept.
\[\left( \frac{4 - 5k}{2k + 1} \right) = 2\left( \frac{4 - 5k}{3 - k} \right)\]
\[ \Rightarrow \left( 4 - 5k \right)\left( 3 - k \right) = \left( 4k + 2 \right)\left( 4 - 5k \right)\]
\[ \Rightarrow \left( 4 - 5k \right)\left( 3 - k - 4k - 2 \right) = 0\]
\[ \Rightarrow \left( 4 - 5k \right)\left( 1 - 5k \right) = 0\]
\[\Rightarrow 4 - 5k = 0 or 1 - 5k = 0\]
\[ \Rightarrow k = \frac{4}{5} or k = \frac{1}{5}\]
When \[k = \frac{4}{5}\] , the equation of the plane is
\[\left( 2 \times \frac{4}{5} + 1 \right)x + \left( \frac{4}{5} + 2 \right)y + \left( 3 - \frac{4}{5} \right)z = 4 - 5 \times \frac{4}{5} \Rightarrow 13x + 14y + 11z = 0\]
This plane does not satisfies the given condition, so this is rejected.
When \[k = \frac{1}{5}\] , the equation of the plane is \[\left( 2 \times \frac{1}{5} + 1 \right)x + \left( \frac{1}{5} + 2 \right)y + \left( 3 - \frac{1}{5} \right)z = 4 - 5 \times \frac{1}{5} \Rightarrow 7x + 11y + 14z = 15\]
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