Let $M$ be the plane $2x + y + 2z = 27$, $S$ the sphere of radius $15$ centered at the origin and let $C$ be the circle of intersection between $M$ and $S$. Determine the points in $M$, $S$ and $C$ that lie closest to the point $P(33, 30, 6)$. (ON-system assumed.)
(1) The closest point from $M$ to $P$ is the intersection of $L_1:P+t_1(2,1,2)$ and $M$, I got this one.
(2) The closest point from $S$ to $P$. Let $F$ be the center of the sphere. Then I think the point of intersection between the sphere and the line that goes through $P$ and $F$ is the desired point.
So $L_2:t_2 (33, 30, 6)$ and we have $S:x^2+y^2+z^2=15^2$. Then $t_2=\frac{15^2}{33^2+30^2+6^2}$, but this seems wrong.
(3) No idea.
(1) Good work!
(2) This seems correct, with the exception of forgetting to take a square root.
(3) Probably possible to bash out with Lagrange multipliers, but I also believe the following geometric argument is correct. To minimize the distance from a point on $M$ to $P$, we simply need to minimize it's distance to the point calculated in part (1) (lets call this point $x_0$). Now, the problem is to find the point on both $M$ and $S$ that is closest to $x_0$. If I am not mistaken in saying that $x_0$ is outside of $C$, then it follows that this point lies between the point in $M$ closest to the origin and $x_0$. So just find the point on the line between $\frac{27}{5}(2,1,2)$ and $x_0$ with norm 15, and you are done.