Sphere Center From Four Points

The Cartesian equation of a sphere with centre $(h,k,l)$ and radius $r$ is given by: $(x-h)^2+(y-k)^2+(z-l)^2=r^2$

Thus, we have: $\begin{cases}(5-h)^2+(8.732-k)^2+(15-l)^2 = r^2\\(2-h)^2+(7-k)^2+(17-l)^2 = r^2\\(3.732-h)^2+(4-k)^2+(15-l)^2 = r^2\\(0.653-h)^2+(10.702-k)^2+(13.695-l)^2 = r^2\end{cases}$

Solving, we obtain $h\approx2,k\approx7,l\approx13$ so the answer is 22.

We know that the distance from a point (x,y,z) to the center of the sphere(h,k,m) is given by $\sqrt{(x-h)^2 + (y-k)^2 + (z-m)^2}$ .

The first thing that pops out of the problem are the numbers 8.732 and 3.732. We know that $\sqrt{3} = 1.732$ . There is always a feeling that certain numbers are given in a problem to make the solution easier.

Assigning h = 2 and k = 7 makes the terms $(x_3 - h)^2$ and $(y_1 - k)^2$ equal to 3.

Now, since all points on a sphere are equidistant from the center of the sphere, we can choose to equate the distances from the points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ with $h = 2$ and $k = 7$

$(x_1 - h)^2 + (y_1 - k)^2 + (z_1 - m)^2 = (x_2 - h)^2 + (y_2 - k)^2 + (z_2 - m)^2$

$(5 - 2)^2 + (8.732 - 7)^2 + (15 - m)^2 = (2 - 2)^2 + (7 - 7)^2 + (17 - m)^2$

$9 + 3 + m^2 - 30m + 225 = 0 + 0 + m^2 - 34m + 289$

Rearranging the terms, we get

$4m = 52$ or $m = 13$

Therefore, $\boxed{h = 2, k = 7, m = 13}$ or $\boxed{h+k+m = 22}$ .