Minimum distance from a plane to an ellipsoid

First, let's analyze what is being asked in the problem. Notice that $x_1^* + y_1^* + z_1^* = 45$ regardless of the values, and that $d^*$ can be obtained from the values of $x_2^*,y_2^*,z_2^*$ by using the formula that computes the distance between a point and a plane. Thus, I will focus only on finding the values of $x_2^*,y_2^*.z_2^*$ . The rest is a trivial computation.

Let's say that the points in the ellipse are $(x,y,z)$ and the points in the plane are $(a,b,c)$ . We want to minimize the squared distance function $F(x,y,z,a,b,c) = (x-a)^2 + (y-b)^2 + (z-c)^2$ . We know from Lagrange Multipliers that extreme values occur at points such that $\nabla F = \mu\nabla (a+b+c-45) + \lambda \nabla \left( \frac{x^2}{10^2} + \frac{y^2}{15^2} + \frac{z^2}{20^2} - 1 \right)$ .

Let's now carefully analyze this. We notice that $F_x = F_a = 2(x-a)$ . And the equation says that $2(x-a) = \lambda \frac{2x}{10^2}$ and $2(x-a) = \mu$ . But we can combine both of these expressions to get $\frac{\mu}{\lambda} = \frac{2x}{10^2}$ . Division by $\lambda$ is valid because if $\lambda = 0$ we would have $2(x-a) = 0 \implies x=a$ . But then we would have also $y = b, z = c$ which would imply an intersection between the surfaces exist. No such intersection exists (but notice that if it did, then this would give the solution with distance 0). So we can conclude that $\frac{\mu}{\lambda} = \frac{2x}{10^2}$ . But what we did with $x,a$ we can also do with $y,b$ and $z,c$ . So what we actually have is that $\frac{\mu}{\lambda} = \frac{2x}{10^2} = \frac{2y}{15^2} = \frac{2z}{20^2}$ .

Using the previous result, with some algebra we can write $\frac{x^2}{10^2} + \frac{y^2}{15^2} + \frac{z^2}{20^2} = x^2 \left ( \frac{1}{10^2} + \frac{15^2}{10^4} + \frac{20^2}{10^4} \right)$ . But this implies that $x^2 \left ( \frac{1}{10^2} + \frac{15^2}{10^4} + \frac{20^2}{10^4} \right) = 1 \implies x^2 = \frac{400}{29} \implies x = \pm \frac{20}{\sqrt{29}}$ . Then $(x_2^*,y_2^*,z_2^*) = \frac{\pm}{\sqrt{29}} \left( 20,45,80 \right)$ .

To find which sign we have to use, just compute the distance for both. If we pick the positive sign, we get $d \approx 10$ while the negative sign gives $d \approx 40$ . Thus the values we want are $\frac{1}{\sqrt{29}} \left( 20,45,80 \right)$ .