Mathematics - 3D-Coordinate Geometry Question with Solution | TestHub

Equation of plane containing line $x-y-1=0$ and $z-1=0$ is $(x-y-1)+\lambda (z-1)=0$

$\Rightarrow x-y+\lambda z-1-\lambda =0$   $...\left(\mathrm{i}\right)$

Since, this plane is parallel to line $\frac{x}{2}-\frac{z}{3}=1$ and $y=3$

$\frac{x-2}{2}=\frac{y-3}{0}=\frac{z}{3}$   $...\left(\mathrm{ii}\right)$

$\therefore 2\left(1\right)+\left(0\right)(-1)+3\left(\lambda \right)=0$

$\Rightarrow \lambda =-2/3$

Putting the value of $\lambda$ in Eq. $\left(\mathrm{i}\right)$, we get

$$\begin{gathered} x-y-\frac{2}{3}z-1+\frac{2}{3}=0 \\ \Rightarrow 3x-3y-2z=1 \end{gathered}$$