not really sure which method to use let there be $A(2,4,6)$ , $B(6,2,2)$; On the $x$ axis, find a point $P$ such that the sum of it's distances from points $A$ and $B$ would be minimal.
I'm not really sure which method I should use here since using derivatives sounds too complex for this and neither vectors nor analytic geometry seem to provide a solution...
 A: This problem is a variant on the famous two-dimensional version. For that version you can either use calculus (as in the OP) or you can use a trick to use geometry. You can use the geometric approach either to avoid or to check the calculus approach.
The 2D geometry trick is to "reflect" one of the two points in the line and note that the path from the un-reflected point to the image of the reflected point is a straight line when the path from point to point via the line has minimum length. So find the intersection of the line segment from the un-reflected point to the image point with the given line to find the desired point.
This three-dimensional version adds a wrinkle, since we cannot just reflect a point in the line. However, we can rotate a point around the line, and if we rotate one point to be opposite the other point, again the solution is the intersection of the line segment from the un-rotated point to the image point with the given line.
In this particular problem, let's view the scene along the $x$-axis, so we see the projection of the points onto the $yz$ plane. We are then viewing the $x$-axis as the origin and our two points are $(4,6)$ and $(2,2)$. Since the direction of $(2,2)$ is obviously $45^\circ$ from the $y$-axis, let's rotate $(4,6)$ to be opposite $(2,2)$. The distance of $(4,6)$ from the origin is clearly $\sqrt{4^2+6^2}=\sqrt{52}$. The point with the same distance from the origin and opposite to $(2,2)$ is $(-\sqrt{26},-\sqrt{26})$.
If we move back to 3D we have rotated the point $A(2,4,6)$ around the point $A0(2,0,0)$ to get the point $A'(2,-\sqrt{26},-\sqrt{26})$. We now see that any path from $B(6,2,2)$ to the $x$-axis to $A(2,4,6)$ has the same length as a path from $B(6,2,2)$ to the $x$-axis to $A'(2,-\sqrt{26},-\sqrt{26})$. But, by construction, the straight line segment path between those two points intersects the $x$-axis and that is obviously the shortest such path.
Therefore we just need to find the intersection $C$ of the line segment from $B(6,2,2)$ to $A'(2,-\sqrt{26},-\sqrt{26})$ with the $x$-axis. There are several ways to do this. One way is with proportions, but I'll use the parameterization of the line segment, which is, for $0\le t\le 1$,
$$C(2+4t,-\sqrt{26}+(2+\sqrt{26})t,-\sqrt{26}+(2+\sqrt{26})t)$$
Solving for $y=0$ or $z=0$ we get $t=\frac{\sqrt{26}}{2+\sqrt{26}}$. Substituting that into the parameterization of $x$ and simplifying, we get
$$x=\frac{74-4\sqrt{26}}{11}$$
and the desired point $C$ is $C(\frac{74-4\sqrt{26}}{11},0,0)$.
This analytic-geometry answer should now be checked with the calculus answer. It is debatable which method is easier, but it certainly is good to use two methods to check each other.


I see that in a comment to another question you ask what to do if the desired point $C$ is said to be in the $yz$ plane (or other plane). The analytic geometry is easier for this problem. If the two points are on the same side of the plane, reflect one of the points in the plane then find the intersection of the line segment from the reflected point to the non-reflected point with the plane.
With your points, reflect point $A(2,4,6)$ in the $yz$ plane to get $A'(-2,4,6)$. A parameterization of the line segment from $A'(-2,4,6)$ to $B(6,2,2)$, for $0\le t\le 1$ is
$$(-2+8t,4-2t,6-4t)$$
Setting the $x$-coordinate to zero and solving gives $t=\frac 14$, so the desired in the $yz$ plane is
$$C(0,3.5,5)$$
This is far easier than the calculus-of-two-variables approach!
To use the $xy$ plane you would reflect $A(2,4,6)$ to $A'(2,4,-6)$. To use the $xz$ plane you would reflect $A(2,4,6)$ to $A'(2,-4,6)$.
Any questions?
A: I assume you know how to write the distance between two points in space.
The point on the x axis is (x,0,0). By using this in the sum of distances you get a function with only x as variable.
Taking the derivative and find x that makes it zero should result the desired x value.
