The point of inter section of the plane  $\overline{r}.\left(3\overline{i}-5\overline{j}+2\overline{k}\right)=6$ and the straight line which is passing through the origin and perpendicular to the plane $2x-y-z=4$is$\left(x_0,y_0,z_0\right)$ then the value of  $\left(2x_0-3y_0+z_0\right)$ is 

Equation of line passing through the point  $\left(x_1,y_1,z_1\right)$and perpendicular to the plane  $ax+by+cz+d=0$is$\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}$

Hence, the equation of the line passing through the origin and perpendicular to the plane  $2x-y-z=4$ is $\frac{x}{2}=\frac{y}{-1}=\frac{z}{-1}$

General point on the above line is $\left(2k,-k,-k\right)$, the given plane is  $3x-5y+2z=6$

If the point lies on the plane then,$6k+5k-2k=6\Rightarrow 9k=6\Rightarrow k=\frac{2}{3}$

Hence, the point of intersection is  $\left(\frac{4}{3},-\frac{2}{3},-\frac{2}{3}\right)$

Consider the expression  $\left(2x_0-3y_0+z_0\right)=\frac{8}{3}+\frac{6}{3}-\frac{2}{3}=\frac{12}{3}=\boxed{4}$