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Let the equation of plane passing through the line of intersection of the planes $\mathrm{x+2y+az=2 \: and\: x-y+z=3}$ be $\mathrm{5x-11y + bz = 6a -1}$ . For $\mathrm{c\: \epsilon \: \: Z}$ , if the distance of this plane from the point $\mathrm{ (a, -c, c) }$ is $\frac{2}{\sqrt{a}}$ , then $\frac{a+b}{c}$ is equal
Option: 1
$-4$

Option: 2
$2$

Option: 3
$-2$

Option: 4
$4$

Answers (1)

best_answer

$\begin{aligned} & (x+2 y+a z-2)+\lambda(x-y+z-3)=0 \\ \end{aligned}$

$\begin{aligned} & \frac{1+\lambda}{5}=\frac{2-\lambda}{-11}=\frac{a+\lambda}{b}=\frac{2+3 \lambda}{6 a-1} \\ \end{aligned}$

$\begin{aligned} & \lambda=-\frac{7}{2}, a=3, b=1 \\ \end{aligned}$

$\begin{aligned} & \frac{2}{\sqrt{a}}=\left|\frac{5 a+11 c+b c-6 a+1}{\sqrt{25+121+1}}\right| \\ \end{aligned}$

$\begin{aligned} & c=-1 \\ \end{aligned}$

$\begin{aligned} & \therefore \frac{a+b}{c}=\frac{3+1}{-1}=-4 \end{aligned}$

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