Equation of the plane through the lines x−12=y+1−2=z−31 andx−11=y+12=z−32  is

# Equation of the plane through the lines $\frac{x-1}{2}=\frac{y+1}{-2}=\frac{z-3}{1}$ and$\frac{x-1}{1}=\frac{y+1}{2}=\frac{z-3}{2}\text{\hspace{0.17em}}$ is

1. A

$2x+y-2z+5=0$

2. B

$2x+y-2z-5=0$

3. C

$x+2y-2z+7=0$

4. D

$x+2y+2z-5=0$

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### Solution:

The Equation of the plane through the lines $\frac{x-1}{2}=\frac{y+1}{-2}=\frac{z-3}{1}$ and$\frac{x-1}{1}=\frac{y+1}{2}=\frac{z-3}{2}\text{\hspace{0.17em}}$ is $\left|\begin{array}{ccc}x-{x}_{1}& y-{y}_{1}& z-{z}_{1}\\ {l}_{1}& {m}_{1}& {n}_{1}\\ {l}_{2}& {m}_{2}& {n}_{2}\end{array}\right|=0$

Hence the equation of the plane required is $\left|\begin{array}{ccc}x-1& y+1& z-3\\ 2& -2& 1\\ 1& 2& 2\end{array}\right|=0$

Simplify

$\begin{array}{c}\left(x-1\right)\left(-4-2\right)-\left(y+1\right)\left(3\right)+\left(z-3\right)\left(6\right)=0\\ 2\left(x-1\right)+\left(y+1\right)-2\left(z-3\right)=0\\ 2x+y-2z+5=0\end{array}$

Therefore, the equation of the required plane is $\overline{)2x+y-2z+5=0}$

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