Suppose you want to maximize $z=f(x,y)$ subject to the constraint $g(x,y)=c$. You've used the method of Lagrange multipliers to have found the maximum $M$ and along the way have computed the Lagrange multiplier $\lambda$. Then $\lambda={dM\over dc}$, i.e. $\lambda$ is the rate of change of the maximum value with respect to $c$.
Said another way, you can think of $\lambda$ as approximately the change in $M$ that results from a one unit change in $c$.
Elaborating further: Optimizing $f(x,y)$ subject to $g(x,y)=c$ via Lagrange multipliers leads to $\nabla f=\lambda g$. Let $L(x,y;\lambda):=f(x,y)+\lambda(c-g(x,y))$. Then the constrained optimization problem can be cast as $\nabla L=0$.
From this perspective, ${\partial L\over \partial c}=\lambda$, i.e. $\lambda$ is the rate of change of the quantity being optimized, $L$, with respect to the constraint value, $c$.
Let us discuss the example you were given. Generally, this optimization method uses the following strategy. Let $f(x,y,z)$ be the function that we are attempting to determine the critical points for, subject to the constraint equation $$g(x,y,z)=k$$ for some $k \in \mathbb{R}$. We solve the following system:
$$\nabla f(x,y,z) = \lambda \nabla g(x,y,z) \\g(x,y,z)=k$$
of four equations and four unknowns (note that $\nabla$ is the gradient function which returns the vector composed of partial derivatives with respect to $x$, $y$, and $z$).
In this case, we have $f(x,y,z)=2x+y-2z$ and $g(x,y,z)=x^2+y^2+z^2=4$ (this is a sphere of radius $2$). Thus, we have the following system of equations:
$$\begin{cases}2 = 2\lambda x \,\,\,\,\,\,(f_x = \lambda g_x) \\ 1 = 2\lambda y \,\,\,\,\,\, (f_y=\lambda g_y)\\ -2 = 2\lambda z \,\,\,\,\,(f_z= \lambda g_z)\\ x^2+y^2+z^2=4\end{cases}$$
There are various ways that you can solve this, but we will solve in the following way. Multiplying the first equation by $yz$, the second equation by $xz$, and the third equation by $xy$ and setting each of these equal to one another, we obtain $$2\lambda xyz = \begin{cases} 2yz \\ xz \\ -2xy \end{cases}$$
So, first we have $x = 2y$ upon dividing $2yz=xz$ by $z \neq 0$. Then we also have $z=-2y$ upon dividing $xz=-2xy$ by $x \neq 0$. Finally, we have $x=-z$ upon dividing $2yz = -2xy$ by $2y \neq 0$. Applying this, we substitute for $x$ and $z$ in terms of $y$ into the fourth equation to get
$$x^2 +y^2 +z^2 =4 \implies 4y^2 + y^2 + 4y^2 = 9y^2 = 4 \implies y = \mp \frac{2}{3}$$
I will let you solve for the other $3$ unknowns (consider each case separately: assume $y = -\frac{2}{3}$ and solve for $x,z,\lambda$ and then assume $y=\frac{2}{3}$ and solve for $x,z,\lambda$). Recall from before that $z = -2y$ and $x=-z$. You will find the two solutions
$$(x,y,z,\lambda)=\left(\mp \frac{4}{3},\mp \frac{2}{3}, \pm \frac{4}{3},\mp \frac{3}{4}\right) .$$
These solutions $(x,y,z)$ are the critical points of the function $f$ under this constraint $g(x,y,z)=4$ and we can use multiple ways to classify them (as, for instance, maximums, minimums, or saddle points).
Best Answer
I think you can get an answer for $\lambda$ from the equation you have, which gives $\lambda_1 = -8$, $\lambda_2 = 17$. Then if you put it in the equation that you found $x$, only by using the equation $x^2+y^2=25$ you can solve that one.