Find the equation of the plane passing through the point (2, –1, 1) and the line $4x-3y+5=0, y-2z-5=0$.

The equation of any plane passing through the line of intersection of the planes $4x-3y+5=0$ and $y-2z-5=0$is $(4x-3y+5)+\lambda (y-2z-5)=0$ which is passing through

$(2, –1, 1)$ so

                         $\lambda =1/2$

Hence, the equation of the required plane is

$\begin{array}{l} (4x-3y+5)+\frac{1}{2}(y-2z-5)=0\\ \Rightarrow 8x-5y-2z+5=0\end{array}$