Find the minimum and maximum distance between the ellipsoid of ecuation $x^2+y^2+2z^2=6$ and the point $P=(4,2,0)$

Taking from where you left off. By equating coordinates of both sides, you have: $2z = 4\lambda z\implies z = 0$ or $\lambda = \dfrac{1}{2}$. If $\lambda = \dfrac{1}{2} \implies x = 8$ which is not possible since $x^2 \le 6$. So what is left is $z = 0$ or $\lambda = 0$. Also $\lambda \neq 1$ for otherwise $2x-8 = 2x$ which is not possible. $\lambda \neq 0$, for if it were $0$, then you have: $2x-8 = 0 \implies x = 4$, and this is not possible since $x^2 \le 6$. Thus $2x-8 = 2\lambda x, 2y-4 = 2\lambda y\implies x = \dfrac{4}{1-\lambda}, y = \dfrac{2}{1-\lambda}\implies \dfrac{16}{(1-\lambda)^2}+\dfrac{4}{(1-\lambda)^2}=6\implies (1-\lambda)^2 = \dfrac{20}{6} = \dfrac{10}{3}\implies \lambda = 1\pm \dfrac{\sqrt{30}}{3}$. Can you take it from here ?. It looks as if one of the lambdas will yield a min and the other a max.

Since by C-S $$2x+y\leq\sqrt{(x^2+y^2)(2^2+1^2)}=\sqrt{5(x^2+y^2)},$$ we obtain $$(x-4)^2+(y-2)^2+z^2=x^2+y^2+z^2-4(2x+y)+20\geq $$ $$\geq x^2+y^2+z^2-4\sqrt{5(x^2+y^2)}+20=6-2z^2+z^2-4\sqrt{5(6-2z^2)}+20=$$ $$=26-z^2-4\sqrt{5(6-2z^2)}\geq26-4\sqrt{30}.$$ It's obvious that the equality occurs, which gives the minimal value: $$\sqrt{26-4\sqrt{30}}.$$

Since $-2x-y\leq\sqrt{5(x^2+y^2)}$, by the same way we can get a maximal value: $$\sqrt{26+4\sqrt{30}}.$$

Done!

hint The ellipse can be parametrized as

$$x_e =\sqrt {6}\sin (\phi)\cos (\theta) $$ $$y_e=\sqrt{6}\sin (\phi)\sin (\theta ) $$ $$z_e=\sqrt {3}\cos (\phi) $$

the square of the distance from a point of the ellipse to the point $(4,2,0) $ is

$$D^2=(x_e-4)^2+(y_e-2)^2+z_e^2$$ $$=3\sin^2 (\phi)+23-8x_e-4y_e $$

You can find min and max D,

by solving the system, $$\frac {\partial D^2}{\partial \phi}=\frac {\partial D^2}{\partial \theta}=0$$