Show that the lines `(x+1)/(-3)=(y-3)/2=(z+2)/1\ a n d\ x/1=(y-7)/(-3)=(z+7)/2` are coplanar. Also, find the equation of the plane containing them.

Explanation

`\frac{x-x_{1}}{\alpha}=\frac{y-y_{1}}{\beta}=\frac{z-z_{1}}{\gamma} \& \frac{x-x_{2}}{a}=\frac{y-y_{2}}{b}=\frac{z-z_{2}}{c}`


`\Rightarrow \quad\left|\begin{array}{ccc}x_{2}-x_{1} & y_{2}-y_{1} & z_{2}-z_{1} \\ \alpha & \beta & \gamma \\ a & b & c\end{array}\right|=0 .`


equation of plane `\Rightarrow \quad\left|\begin{array}{ccc}x-x_{1} & y-y_{1} & z-z_{1} \\ \alpha & \beta & \gamma\\ a & b & c\end{array}\right|=0`


`\frac{x+1}{-3}=\frac{y-3}{2}=\frac{z+2}{1} \& \quad \frac{x}{1}=\frac{y-7}{-3}=\frac{z+7}{2}`.


`x_{1}=-1, y_{1}=3 \quad z_{1}=-2 . \quad x_{2}=0 \quad y_{2}=7 \quad z_{2}=-7 .`


`\alpha=-3 \quad \beta=2, \gamma=1 \quad a=1, b=-3, c=2`.


For coplanarity, `\left|\begin{array}{ccc}1 & 4 & -5 \\ -3 & 2 & 1 \\ 1 & -3 & 2\end{array}\right|`


`=1(4+3)-4(-6-1)-5(9-2)`


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