Question

# The equation of the plane through the line x + y + z + 3 = 0 = 2x − y + 3z + 1 and parallel to the line $\frac{x}{1}=\frac{y}{2}=\frac{z}{3}$ is (a) x − 5y + 3z = 7 (b) x − 5y + 3z = −7 (c) x + 5y + 3z = 7 (d) x + 5y + 3z = −7

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Solution

## (a) x − 5y + 3z = 7 $\text{The equation of the plane passing through the line of intersection of the given planes is}\phantom{\rule{0ex}{0ex}}x+y+z+3+\lambda \left(2x-y+3z+1\right)=0\phantom{\rule{0ex}{0ex}}\left(1+2\lambda \right)x+\left(1-\lambda \right)y+\left(1+3\lambda \right)z+3+\lambda =0...\left(1\right)\phantom{\rule{0ex}{0ex}}\text{This plane is parallel to the line}\frac{x}{1}\text{=}\frac{y}{2}\text{=}\frac{z}{3}.\text{It means that this line is perpendicular to the normal of the plane (1).}\phantom{\rule{0ex}{0ex}}⇒1\left(1+2\lambda \right)+2\left(1-\lambda \right)+3\left(1+3\lambda \right)=0\text{(Because}{a}_{1}{a}_{2}+{b}_{1}{b}_{2}+{c}_{1}{c}_{2}=0\text{)}\phantom{\rule{0ex}{0ex}}⇒1+2\lambda +2-2\lambda +3+9\lambda =0\phantom{\rule{0ex}{0ex}}⇒9\lambda +6=0\phantom{\rule{0ex}{0ex}}⇒\lambda =\frac{-2}{3}\phantom{\rule{0ex}{0ex}}\text{Substituting this in (1), we get}\phantom{\rule{0ex}{0ex}}\left(1+2\left(\frac{-2}{3}\right)\right)x+\left(1-\left(\frac{-2}{3}\right)\right)y+\left(1+3\left(\frac{-2}{3}\right)\right)z+3+\left(\frac{-2}{3}\right)=0\phantom{\rule{0ex}{0ex}}⇒-x+5y-3z+7=0\phantom{\rule{0ex}{0ex}}⇒x-5y+3z=7$

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