Extreme Tangents

A normal to the ellipsoid $f(x,y,z) = \tfrac{1}{100}x^2 + \tfrac{1}{25}y^2 + \tfrac{1}{!69}z^2 = 1$ at the point $\mathbf{r} \; = \; \left(\begin{array}{c} x \\ y \\ z \end{array}\right)$ is $\mathbf{n} \; = \; \tfrac12\nabla f \; = \; \left(\begin{array}{c} \tfrac{1}{100}x \\ \tfrac{1}{25}y \\ \tfrac{1}{169}z \end{array} \right)$ and so the condition for the point $\mathbf{r}$ to be the position vector of an end of a tangent segment is $\mathbf{n} \cdot \left[ \left(\begin{array}{c} 7 \\ 11 \\ 20 \end{array}\right) - \mathbf{r}\right] \; = \; 0$ and hence that $\tfrac{7}{100}x + \tfrac{11}{25}y + \tfrac{20}{169}z \; = \; 1$ Thus we need to maximize and minimize $\sqrt{(x-7)^2 + (y-11)^2 + (z-20)^2}$ subject to the pair of constraints $\tfrac{1}{100}x^2 + \tfrac{1}{25}y^2 + \tfrac{1}{169}z^2 \; =\; \tfrac{7}{100}x + \tfrac{11}{25}y + \tfrac{20}{169}z \; = \; 1$ Solving these problems numerically, we see that the maximum value is $M = 28.8461$ , occurring at $(0.889242,4.10408,-7.33496)$ , while the minimum value is $m = 15.3581$ , occurring at $(2.90981,-1.39088,11.9001)$ .

Thus the answer is $M + m = \boxed{44.204}$ to $3$ decimal places.