Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

$\left|\begin{array}{ccccccc}1-2& 0+1& -1-01& -1& 11& 2& 3\end{array}\right|=0\Rightarrow$ given lines are coplanar. Equation of the plane is $\left|\begin{array}{ccccccc}x-1& y& z+11& -1& 11& 2& 3\end{array}\right|=0$, i.e. $5x+2y-3z-8=0$. Since $\frac{1}{3}=\frac{2}{6}=\frac{3}{9}\Rightarrow \frac{x-2}{1}=\frac{y+1}{2}=\frac{z}{3}$ is perpendicular to the plane $3x+6y+9z-8=0$, and also $1(1)+2(1)+3(-1)=0\Rightarrow$ the line is parallel to $x+y-z=0$.