Find a unit vector $n$ that is normal to the surface $2 z^{2}-x^{4}-4 y^{4}=30$ at $P=(2,1,5)$ that points in the direction of the $x y$ plane (in other words, if you travel in the direction of $\mathbf{n}$, you will eventually cross the $x y$-plane).

Use symbolic notation and fractions where needed.)

Step 1 ：The function that defines the surface is \(f(x, y, z) = 2z^2 - x^4 - 4y^4 - 30\).

Step 2 ：The gradient of \(f\) is given by the vector \(\nabla f = \langle f_x, f_y, f_z \rangle\), where \(f_x\), \(f_y\), and \(f_z\) are the partial derivatives of \(f\) with respect to \(x\), \(y\), and \(z\) respectively.

Step 3 ：We have \(f_x = -4x^3\), \(f_y = -16y^3\), and \(f_z = 4z\).

Step 4 ：At the point \(P=(2,1,5)\), we have \(f_x(P) = -4(2)^3 = -32\), \(f_y(P) = -16(1)^3 = -16\), and \(f_z(P) = 4(5) = 20\).

Step 5 ：So, \(\nabla f(P) = \langle -32, -16, 20 \rangle\).

Step 6 ：This is a vector normal to the surface at \(P\). However, we want a unit vector that points in the direction of the \(xy\)-plane. The \(xy\)-plane is defined by \(z=0\), so we want the \(z\)-component of our vector to be negative. Therefore, we take \(-\nabla f(P)\) as our normal vector.

Step 7 ：\(-\nabla f(P) = \langle 32, 16, -20 \rangle\).

Step 8 ：To make this a unit vector, we divide by its magnitude. The magnitude of a vector \(\langle a, b, c \rangle\) is given by \(\sqrt{a^2 + b^2 + c^2}\).

Step 9 ：The magnitude of \(-\nabla f(P)\) is \(\sqrt{(32)^2 + (16)^2 + (-20)^2} = \sqrt{1024 + 256 + 400} = \sqrt{1680}\).

Step 10 ：So, the unit vector normal to the surface at \(P\) that points in the direction of the \(xy\)-plane is \(\frac{1}{\sqrt{1680}}\langle 32, 16, -20 \rangle = \langle \frac{32}{\sqrt{1680}}, \frac{16}{\sqrt{1680}}, -\frac{20}{\sqrt{1680}} \rangle\).

Step 11 ：\(\boxed{\langle \frac{32}{\sqrt{1680}}, \frac{16}{\sqrt{1680}}, -\frac{20}{\sqrt{1680}} \rangle}\) is the final answer.