Any general point on the line is $\left(2\lambda ,5\lambda +\mathrm{1,3}\lambda -1\right)$

On satisfying this point on the plane, we get, 
$2\lambda +5\lambda +1+6\lambda -2=3$
$13\lambda =4\Rightarrow \lambda =\frac{4}{13}$
So, coordinates of the point are $\left(\frac{8}{13},\frac{33}{13},\frac{-1}{13}\right)$
This point also lies on the image of the line
Image of point $\left(\mathrm{0,1},-1\right)$ also lies on the image of the line
$\frac{x-0}{1}=\frac{y-1}{1}=\frac{z+1}{2}=-2\frac{\left(-4\right)}{6}$
$x=\frac{4}{3},y=\frac{7}{3},z=\frac{5}{3}$
Point is $\left(\frac{4}{3},\frac{7}{3},\frac{5}{3}\right)$
Equation of image of the line is $\frac{x-\frac{4}{3}}{28}=\frac{y-\frac{7}{3}}{-8}=\frac{z-\frac{5}{3}}{68}$
For $xz-$plane, putting $y=0$, we get, 

$\frac{x-\frac{4}{3}}{28}=\frac{7}{24}=\frac{z-\frac{5}{3}}{68}$
$\Rightarrow z=\frac{129}{6}$