Finding a tangent plane equation

Question

I have to find the tangent plane equation to the surface $zx^2+xy^2+yz^2=5$ at the point of $(-1,1,2)$.

I couldn't get the right answer.

Answer 1

Let $f(x,y,z) = zx^2 +xy^2 +yz^2 $. The normal vector of the curve at the point $(-1,1,2)$ is

$$(f_x,f_y,f_z)=(2xz+y^2,2xy+z^2,2yz+x^2) =(-3,2,5)$$

Then, the equation of the plane is given by

$$(x+1,y-1,z-2)\cdot (-3,2,5) =0$$