Using Lagrange Multipliers to determine the point on a surface nearest to P I'm attempting to figure this problem out. I would appreciate some guidance on how to get the answer. Thanks.
Consider the surface defined as $S: x^2+y^2+z^2 = 8$. If we have a $P = (0,1,1)$, use Lagrange multipliers to determine the point on $S$ nearest to $P$.

It is obvious that the  shortest distance between $P$ and $S$ is the distance between $P$ and the  center of the sphere, minus the radius, that is:
$$
|\sqrt{0^2+1^2+1^2}-\sqrt{8}| = \sqrt{2}
$$
Now, if you want the coordinates of the nearest point, you can use Lagrange multipliers indeed: you want to minimize the (squared) distance between $P$ and $S$, given by
$$
x^2+(y-1)^2+(z-1)^2
$$
subject to 
$$
x^2+y^2+z^2=8
$$
The Lagrangian equals 
$$
\mathcal{L}=x^2+(y-1)^2+(z-1)^2 + \lambda(8-x^2+y^2+z^2)
$$
So you need to solve the following system
\begin{cases}
2x-2\lambda x =0 \\
2y-2-2\lambda y = 0 \\
2z -2 -2\lambda z = 0 \\
x^2+y^2+z^2=8
\end{cases}
You should get
$$
(x,y,z,\lambda)=(0,2,2,\frac{1}{2})
$$
So the nearest the point has coordinates $(0,2,2)$ and the shortest distance equals 
$$\sqrt{0^2+(2-1)^2+(2-1)^2}=\sqrt{2}$$
