Find an equation for the tangent plane to the surface $2xz^2+3xy-4x=7$ at the point $(1,-1,2)$, by considering a gradient of a suitable scalar field. This is what I've done, followed by a few questions I don't undestand;
$f(x, y, z) = 2xz^2-3xy-4x=7$
Surface is at 0, So therefore we set our equation to zero $2xz^2-3xy-4x-7=0$
We then take partials and evaluate them at the point.
$F_x = x-2, F_x(1,-1,2) = 1-2=3$
$F_y = 1, F_y(1,-1,2) = 1$
$F_z = 2z, F_z(1,-1,2) = 2(2) = 4$
Now the tangent plane is $z=f(x_0,y_0)+f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0)$
=$-3(x-1)+1(y+1)+4(z-1) = 0$
=$-3x+y+4z-6=0$
My questions;

*

*I found this method of computing online and it was done for a vector. Now given this is a scalar field would it still work?


*In the end we have $-3x+y+4z-6=0$, what exactly do I need to solve for? Should I bring the equation back in the original form where it was equal to a number? Or do I solve for z, x or y?


*Why is surface at zero and why do we take partials?
 A: Consider the function $F(x, y, z) = 2xz^2-3xy-4x-7.$ The level surfaces of $F$ are perpendicular to the vector field $\nabla F=(F_x,F_y,F_z).$ Thus, the equation of the tangent plane of a level surface at $(x_0,y_0,z_0)$ is given by $$F_x(x_0,y_0,z_0)(x-x_0)+F_y(x_0,y_0,z_0)(y-y_0)+F_z(x_0,y_0,z_0)(z-z_0)=0.$$ In your case, it is $$7(x-1)-3(y+1)+8(z-2)=0.$$
In the particular case where we have $z=f(x,y)$ then the tangent plane at $(x_0,y_0)$ is given by $z=f(x_0,y_0)+f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0).$ Just consider $F(x,y,z)=z-f(x,y)$ and apply what was said in the paragraph above.
A: I think you made a mistake in finding the gradient vector. From there on, given you have to find the scalar equation of the plane, you can do this way -
$f(x, y, z) = 2xz^2-3xy-4x-7 = 0$ (as called out in comments, the function in the title seems wrong.
$f_x = 2z^2 - 3y - 4$
$f_y = -3x$
$f_z = 4xz$
As the gradient vector is orthogonal to the surface at a given point, it is also orthogonal (normal) to the tangent plane at that point.
Then at point $(1, -1, 2)$, normal vector to the plane will be $(7, -3, 8)$
The general equation of the plane or the scalar equation is given by,
$a(x-x_0) + b(y-y_0) + c(z-z_0) = 0$
where $(a, b, c)$ is the normal vector to the plane and $(x_0, y_0, z_0)$ is a point on the plane.
So the equation of the plane is -
$7(x-1) - 3(y+1) + 8(z-2) = 0$
i.e. $7x - 3y + 8z = 26$
