Answer
The parametric equations of the tangent line to $C$ at $P = \left( {1,1,1} \right)$ is
${\bf{r}}\left( t \right) = \left( {1 + 8t,1 - 8t,1} \right)$ ${\ \ }$ for $ - \infty < t < \infty $
Work Step by Step
We are given two spheres ${x^2} + {y^2} + {z^2} = 3$ and ${\left( {x - 2} \right)^2} + {\left( {y - 2} \right)^2} + {z^2} = 3$.
Let $F\left( {x,y,z} \right) \equiv {x^2} + {y^2} + {z^2}$ and $G\left( {x,y,z} \right) \equiv {\left( {x - 2} \right)^2} + {\left( {y - 2} \right)^2} + {z^2}$.
Thus, the two spheres can be written as $F\left( {x,y,z} \right) \equiv {x^2} + {y^2} + {z^2} = 3$ and $G\left( {x,y,z} \right) \equiv {\left( {x - 2} \right)^2} + {\left( {y - 2} \right)^2} + {z^2} = 3$, respectively.
By Theorem 5, the normal vectors are
$\nabla F = \left( {2x,2y,2z} \right)$, ${\ \ }$ $\nabla G = \left( {2x - 4,2y - 4,2z} \right)$
At the point $P = \left( {1,1,1} \right)$, we have
$\nabla {F_P} = \left( {2,2,2} \right)$, ${\ \ }$ $\nabla {G_P} = \left( { - 2, - 2,2} \right)$
Using the result of Exercise 61, the direction vector of the tangent line to $C$ at $P$ is given by
${\bf{v}} = \nabla {F_P} \times \nabla {G_P}$
${\bf{v}} = \left| {\begin{array}{*{20}{c}}
{\bf{i}}&{\bf{j}}&{\bf{k}}\\
2&2&2\\
{ - 2}&{ - 2}&2
\end{array}} \right| = 8{\bf{i}} - 8{\bf{j}}$
${\bf{v}} = \left( {8, - 8,0} \right)$
Thus, the parametric equations of the tangent line to $C$ at $P = \left( {1,1,1} \right)$ is
${\bf{r}}\left( t \right) = \left( {1,1,1} \right) + \left( {8, - 8,0} \right)t$, ${\ \ }$ for $ - \infty < t < \infty $
${\bf{r}}\left( t \right) = \left( {1 + 8t,1 - 8t,1} \right)$
