Solution (1) We begin by finding the values of the parameters at the point $( -3,1,4) .$ That is, we solve the system of equations $$\left\{ \begin{array}{rcl@{\qquad}r} 3u &=&-3 & \color{#0066A7}{{\hbox{\((1)\)}}}\\ 6-u^{2}-v^{2}&=& 1 & \color{#0066A7}{{\hbox{\((2)\)}}}\\ 2v&=& 4 & \color{#0066A7}{{\hbox{\((3)\)}}} \end{array} \right.$$

From $( 1)$ we find $u=-1,$ and from (3) we find $v=2.$ [Checking, we find these values also satisfy (2).]

We now find the tangent vectors $\mathbf{r}_{u}$ and $\mathbf{r}_{v}.$ $$\begin{array}{@{\hspace*{-2.2pc}}rcl@{\qquad}lrll} \mathbf{r}_{u} &=&\dfrac{\partial }{\partial u}( 3u) \mathbf{i}+ \dfrac{\partial }{\partial u}( 6-u^{2}-v^{2}) \mathbf{j}+\dfrac{ \partial }{\partial u}\left( 2v\right) \mathbf{k}=3\mathbf{i}-2u\,\mathbf{j} & \mathbf{r}_{u}( -1,2) &=&3\mathbf{i}+2\,\mathbf{j}\\[4pt] \mathbf{r}_{v} &=&\dfrac{\partial }{\partial v}( 3u) \mathbf{i}+ \dfrac{\partial }{\partial v}( 6-u^{2}-v^{2}) \mathbf{j}+\dfrac{ \partial }{\partial v}( 2v) \mathbf{k}=-2v\mathbf{j}+2\,\mathbf{k} & \mathbf{r}_{v}( -1,2) &=&-4\mathbf{j}+2\, \mathbf{k} \end{array}$$

Figure 48 The tangent plane to $S$ at $(-3,1,4)$.

The normal vector $\mathbf{n}$ to the tangent plane at $( -3,1,4)$ is $$\mathbf{n} = \mathbf{r}_u ( -1,2) \times \mathbf{r}_v ( -1,2) =\left\vert \begin{array}{c@{\quad}c@{\quad}c} \mathbf{i} & \mathbf{j} & \mathbf{k} \\[3pt] 3 & 2 & 0 \\[3pt] 0 & -4 & 2 \end{array} \right\vert =4\mathbf{i}-6\mathbf{j}-12\mathbf{k}$$

An equation of the tangent plane is $$4( x+3) -6( y-1) -12( z-4) =0\qquad\hbox{or equivalently}\qquad 4x-6y-12z=-66$$

(b) The normal line to the tangent plane contains the point $( -3,1,4)$ and is parallel to the vector $\mathbf{n}=4\mathbf{i}-6\mathbf{j} -12\mathbf{k}$. A parametrization of the normal line is $$\mathbf{r}(t) = ( -3+4t ) \mathbf{i}+ ( 1-6t ) \mathbf{j} + ( 4-12t ) \mathbf{k}$$

See Figure 48.