After applying a rigid motion to $S$ and $P$, we may as well assume $(x_0,y_0,z_0)=(0,0,0)$ and $P$ is the $xy$-plane. Let $U$ be a small neighborhood of the origin in $S$; by taking $U$ to be small enough, we may assume that $U$ is the image of an open disk in $\mathbb R^2$ under a local parametrization, and thus $U$ is homeomorphic to a disk. The hypothesis implies that the origin is the only point of $U$ for which the $z$-coordinate is zero, so $U\smallsetminus\{(0,0,0)\}$ is contained in the set where $z\ne 0 $. Since a disk minus a point is connected, it follows that either $z>0$ on all of $U\smallsetminus\{(0,0,0)\}$ or $z<0$ on all of $U\smallsetminus\{(0,0,0)\}$. After reflecting in the $xy$-plane, we may assume it is $z>0$.
Now suppose $v$ is a tangent vector to $S$ at the origin. Then there is a smooth curve $c:(-\varepsilon,\varepsilon)\to U$ satisfying $c(0) = (0,0,0)$ and $c'(0) = v$. If we write $c(t) = (x(t),y(t),z(t))$, we see that $z(t)$ attains a minimum at $t=0$, and therefore $z'(0)=0$. This means that every tangent vector at the origin has zero $z$-coordinate, so $T_{(0,0,0)}S$ is contained in the $xy$-plane (i.e., in $P$). Since both $T_{(0,0,0)}S$ and $P$ are two-dimensional, they must be equal.
Clearly one direction of the line is given by ${\bf n} = (3,2,-\sqrt{2})$. If the plane is perpendicular to the line, then the normal direction to the plane, the gradient, must be parallel to our $\bf n$.
You want $(x_0,y_0,z_0)$ in the surface such that $\nabla f (x_0, y_0, z_0) = \lambda {\bf n}$ for some $\lambda \in \Bbb R$. You want to solve $$\begin{cases} 3x_0^2 = 3\lambda \\ -4y_0 = 2\lambda \\ 2z_0 = -\lambda \sqrt{2} \\ x_0^3 - 2y_0^2 + z_0^2 = 27 \end{cases}$$ Solve for $x_0, y_0 $ and $z_0$ in terms of $\lambda $ in the first three equations. Substitute in the fourth. Find $\lambda $. Go back and find $x_0, y_0$ and $z_0$. Now find the tangent plane at this point.
(It is possible that more than one point verify these relations, so that the problem might have more than one solution.)
Best Answer
Let $f : U\subset \mathbb{R}^3\to \mathbb{R}$ be given by $f(x,y,z)=\cos x\sin y e^z$ then we have that your surface is indeed a level set $M = f^{-1}(0)$. Then it's easy: remember that the gradient of a function is orthogonal to the level sets. Using this we have that
$$\nabla f(x,y,z)=(-\sin x\sin ye^z, \cos x\cos y e^z, \cos x\sin y e^{z})$$
So that at $(\pi/2,1,0)$ we have $\nabla f(\pi/2,1,0)=(-\sin 1,0,0)$, so that the normal is a multiple of the vector $e_1$, and hence since the magnitude of the normal vector doesn't matter, we can pick the normal vector to be $e_1$. Of course then, the tangent plane is just the $yz$ plane.