5. (20 pts) Find an equation of the tangent plane to the given parametric surface at the specified point.
\[
\vec{r}(u, v)=3 u \vec{\imath}+\left(6-u^{2}-v^{2}\right) \vec{\jmath}+2 v \vec{k}
\]
@ $(-3,1,4)$

Step 1 ：The given parametric surface is \(\vec{r}(u, v)=3 u \vec{i}+(6-u^{2}-v^{2}) \vec{j}+2 v \vec{k}\). We are asked to find the equation of the tangent plane at the point \((-3,1,4)\).

Step 2 ：The equation of the tangent plane to a surface at a given point can be found using the gradient of the surface at that point. The gradient is a vector that points in the direction of the steepest ascent on the surface, and its components are the partial derivatives of the surface equation with respect to each variable.

Step 3 ：First, we need to find the partial derivatives of the surface equation with respect to u and v. These will give us two vectors that are tangent to the surface at the given point. The partial derivatives are \(\vec{r}_u = \begin{bmatrix} 3 \ -2u \ 0 \end{bmatrix}\) and \(\vec{r}_v = \begin{bmatrix} 0 \ -2v \ 2 \end{bmatrix}\).

Step 4 ：At the given point \((-3,1,4)\), these become \(\vec{r}_u = \begin{bmatrix} 3 \ 2 \ 0 \end{bmatrix}\) and \(\vec{r}_v = \begin{bmatrix} 0 \ -4 \ 2 \end{bmatrix}\).

Step 5 ：The cross product of these vectors will give us the normal vector to the surface at that point, which is \(\vec{n} = \begin{bmatrix} 4 \ -6 \ -12 \end{bmatrix}\).

Step 6 ：The equation of the tangent plane is then given by the dot product of the normal vector and the vector from the point on the surface to a general point on the plane being equal to zero. This gives us the equation \(4x - 6y - 12z + 66 = 0\).

Step 7 ：Final Answer: The equation of the tangent plane to the given parametric surface at the point \((-3,1,4)\) is \(\boxed{4x - 6y - 12z + 66 = 0}\).