Answer
The parametric equations of the tangent line to $C$ at $P = \left( {1,2,1} \right)$ is
${\bf{r}}\left( t \right) = \left( {1 - 4t,2 + 26t,1 - 25t} \right)$ ${\ \ }$ for $ - \infty < t < \infty $
Work Step by Step
We are given two surfaces ${x^3} + 2xy + yz = 7$ and $3{x^2} - yz = 1$.
Let $F\left( {x,y,z} \right) \equiv {x^3} + 2xy + yz$ and $G\left( {x,y,z} \right) \equiv 3{x^2} - yz$.
Thus, the two surfaces can be written as $F\left( {x,y,z} \right) \equiv {x^3} + 2xy + yz = 7$ and $G\left( {x,y,z} \right) \equiv 3{x^2} - yz = 1$, respectively.
By Theorem 5, the normal vectors are
$\nabla F = \left( {3{x^2} + 2y,2x + z,y} \right)$, ${\ \ }$ $\nabla G = \left( {6x, - z, - y} \right)$
At the point $P = \left( {1,2,1} \right)$, we have
$\nabla {F_P} = \left( {7,3,2} \right)$, ${\ \ \ }$ $\nabla {G_P} = \left( {6, - 1, - 2} \right)$
Using the result of Exercise 61, the direction vector of the tangent line to $C$ at $P$ is given by
${\bf{v}} = \nabla {F_P} \times \nabla {G_P}$
${\bf{v}} = \left| {\begin{array}{*{20}{c}}
{\bf{i}}&{\bf{j}}&{\bf{k}}\\
7&3&2\\
6&{ - 1}&{ - 2}
\end{array}} \right| = - 4{\bf{i}} + 26{\bf{j}} - 25{\bf{k}}$
${\bf{v}} = \left( { - 4,26, - 25} \right)$
Thus, the parametric equations of the tangent line to $C$ at $P = \left( {1,2,1} \right)$ is
${\bf{r}}\left( t \right) = \left( {1,2,1} \right) + \left( { - 4,26, - 25} \right)t$, ${\ \ }$ for $ - \infty < t < \infty $
${\bf{r}}\left( t \right) = \left( {1 - 4t,2 + 26t,1 - 25t} \right)$
