If the equation of a plane P, passing
Question:

If the equation of a plane $P$, passing through the intersection of the planes, $x+4 y-z+7=0$ and $3 x+y+5 z=8$ is $a x+b y$

Solution:

Equation of plane $P$ is

$(x+4 y-z+7)+\lambda(3 x+y+5 z-8)=0$

$\Rightarrow x(1+3 \lambda)+y(4+\lambda)+z(-1+5 \lambda)+(7-8 \lambda)=0$

$\Rightarrow \frac{1+3 \lambda}{a}=\frac{4+\lambda}{b}=\frac{5 \lambda-1}{6}=\frac{7-8 \lambda}{-15}$

From last two ratios, $\lambda=-1$

$\Rightarrow \frac{-2}{a}=\frac{3}{b}=-1$

$\therefore a=2, b=-3$

$\therefore$ Equation of plane is, $2 x-3 y+6 z-15=0$

Distance $=\frac{|6-6-6-15|}{7}=\frac{21}{7}=3$.

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