Find the equation of the tangent-plane to the surface given with the equation $$2x^2+5y^2+2z^2-2xy+6yz-4x-y-2z=0$$ that passes through the line $$4x-5y=0, \ \ z-1=0.$$
The equation of the slope-plane at the point $M(x_0,y_0,z_0)$ is given by: $$(2x_0-y_0-2)x+(-x_0+5y_0+3z_0-\frac{1}{2})y+(3y_0+2z_0-1)z-2x_0-\frac{1}{2}y_0-z_0=0. \ \ \ \ \ \ (1)$$ (just plugged the coefficients into the general formula of tangent-plane).
Also, as the plan passes through the line given, we obtain it is of the form $$4x-5y+t(z-1)=0 \ \ \text{ or } \ \ 4x-5y+tz-t=0 . \ \ \ \ (2)$$ but the system (with respect to $x_0,y_0,z_0,t$) obtained by equalizing the coefficients of $(1)$ and $(2)$ is inconsistent (has no solution). I got stuck here. Any help is appreciated.
The given line is tangent to the surface at point $P=(0,0,1)$. It suffices then to find the equation of the plane tangent to the surface at $P$, that is: $-4x+5y+2z=2$.