So the tangent plane would be $$T(x_0) = f(x_0) + \nabla f(x-x_0) $$I think the notation here might not be correct for the tangent plane, but I think that I have the right idea (as you see later)
First I find the gradient vector, $\nabla f = (2xy + 3z^2, x^2 + 4yz, 2y^2 + 6zx)$. Then I find $\nabla f (1,1,1) = (5,5,8)$.
Then plugging into the tangent plane equation, I get $$ T(1,1,1) = 6 + (5,5,8) \cdot \begin{pmatrix} x-1 \\ y-1 \\ z-1\end{pmatrix}$$ which simplifies to $T(x_0) = 5x + 5y + 8z - 12$ or $12 = 5x + 5y + 8z$. Is something off with this? Or do the steps look all correct for the most part?
The tangent plane is
$$ T(1,1,1) = 0 + (5,5,8) \cdot \begin{pmatrix} x-1 \\ y-1 \\ z-1\end{pmatrix}$$
which simplifies to $18 = 5x + 5y + 8z$