Question 10 of 10 Let $C$ be the curve of intersection of the spheres $x^{2}+y^{2}+z^{2}=27$ and $(x-2)^{2}+(y-2)^{2}+z^{2}=27$. Find the parametric equations of the tangent line to $C$ at $P=(1,1,5)$. It is known that if the intersection of two surfaces $F(x, y, z)=0$ and $G(x, y, z)=0$ is a curve $C$ and $P$ is a point on $C$, then the vector $\mathbf{v}=\nabla F_{P} \times \nabla G_{P}$ is a direction vector for the tangent line to $C$ at $P$. (Use symbolic notation and fractions where needed. Enter your answers as functions of parameter $t$ in a form $\mathbf{r}(t)=\langle x(t), y(t), z(t)\rangle=\mathbf{r}_{0}+\mathbf{v} t$, where $\mathbf{r}_{0}$ is the corresponding coordinate of point $P$.)

Step 1 ：First, we need to find the gradient vectors of the two given spheres at point P. The gradient of a function F(x, y, z) is given by \(\nabla F = \left(\frac{dF}{dx}, \frac{dF}{dy}, \frac{dF}{dz}\right)\).

Step 2 ：For the first sphere, F(x, y, z) = x^2 + y^2 + z^2 - 27, we have: \(\nabla F_P = \left(\frac{dF}{dx}, \frac{dF}{dy}, \frac{dF}{dz}\right) = (2x, 2y, 2z)\) evaluated at P = (1, 1, 5) gives \(\nabla F_P = (2, 2, 10)\).

Step 3 ：For the second sphere, G(x, y, z) = (x-2)^2 + (y-2)^2 + z^2 - 27, we have: \(\nabla G_P = \left(\frac{dG}{dx}, \frac{dG}{dy}, \frac{dG}{dz}\right) = (2(x-2), 2(y-2), 2z)\) evaluated at P = (1, 1, 5) gives \(\nabla G_P = (-2, -2, 10)\).

Step 4 ：Next, we find the cross product of these two gradient vectors, which will give us the direction vector of the tangent line to the curve C at point P. \(\nabla F_P \times \nabla G_P = (2, 2, 10) \times (-2, -2, 10) = (0, 0, 0)\).

Step 5 ：This result indicates that the two gradient vectors are parallel, which means the two spheres are tangent at point P. Therefore, the tangent line to the curve C at point P does not exist. \(\boxed{\text{The tangent line does not exist}}\)