Solve the following system of equations:

$$
x+y+z=11
$$

$$
x^{2}+2 y^{2}+3 z^{2}=66 \text {. }
$$
I. solution. From equation (1), express $x$ and substitute it into (2):

$$
\begin{aligned}
x & =11-y-z \\
(11-y-z)^{2}+2 y^{2}+3 z^{2} & =66 \\
3 y^{2}+2 y z+4 z^{2}-22 y-22 z+55 & =0 .
\end{aligned}
$$

Consider this as a quadratic equation in $y$:

$$
3 y^{2}+(2 z-22) y+\left(4 z^{2}-22 z+55\right)=0
$$

This has a solution if and only if its discriminant is non-negative:

$$
\begin{aligned}
(2 z-22)^{2}-4 \cdot 3 \cdot\left(4 z^{2}-22 z+55\right) & \geq 0 \\
-44 z^{2}+176 z-176 & \geq 0 \\
-z^{2}+4 z-4 & \geq 0 \\
-(z-2)^{2} & \geq 0
\end{aligned}
$$

Since $(z-2)^{2} \geq 0$, this holds if and only if $(z-2)^{2}=0$, i.e., $z=2$. Then, from (5),

$$
3 y^{2}-18 y+27=0
$$

which has the unique solution $y=3$. Finally, $x=11-y-z=6$.

These numbers satisfy the system of equations.

Remark. The solution can also be stated by transforming the left-hand side of (4) into a sum of squares:

$$
3\left(y+\frac{1}{3} z-\frac{11}{3}\right)^{2}+\frac{11}{3}(z-2)^{2}=0
$$

This holds if and only if both parentheses are zero:

$$
y+\frac{1}{3} z-\frac{11}{3}=0 \quad \text { and } z-2=0
$$

i.e., $z=2$ and $y=3$.

Since the quadratic formula applied to (5) is based on the same idea, these two solutions are essentially the same.

II. solution. Write the weighted arithmetic and quadratic mean inequality for the numbers $x, 2 y, 3 z$ with weights $1, \frac{1}{2}, \frac{1}{3}$ (This inequality holds for any real numbers):

$$
\begin{aligned}
\frac{1 \cdot x+\frac{1}{2} \cdot 2 y+\frac{1}{3} \cdot 3 z}{1+\frac{1}{2}+\frac{1}{3}} & \leq \sqrt{\frac{1 \cdot x^{2}+\frac{1}{2} \cdot(2 y)^{2}+\frac{1}{3} \cdot(3 y)^{2}}{1+\frac{1}{2}+\frac{1}{3}}} \\
\frac{x+y+z}{\frac{11}{6}} & \leq \sqrt{\frac{x^{2}+2 y^{2}+3 y^{2}}{\frac{11}{6}}}
\end{aligned}
$$

From (1) and (2),

$$
6 \leq \sqrt{36}
$$

We see that the two means are equal. It is known that the arithmetic and quadratic means are equal if and only if the numbers in the means are non-negative and equal, i.e.,

$$
x=2 y=3 z>0
$$

From this, $y=\frac{x}{2}$ and $z=\frac{x}{3}$, substituting into (1):

$$
\begin{aligned}
x+\frac{x}{2}+\frac{x}{3} & =11 \\
x & =6
\end{aligned}
$$

$$
y=\frac{x}{2}=3 \quad \text { and } \quad z=\frac{x}{3}=2
$$

Remark. We could also have written the Cauchy-Schwarz-Bunyakovsky inequality for the numbers $x, \sqrt{2} y, \sqrt{3} z$ and $1, \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{3}}$:

$$
\begin{gathered}
\left(x \cdot 1+\sqrt{2} y \cdot \frac{1}{\sqrt{2}}+\sqrt{3} z \cdot \frac{1}{\sqrt{3}}\right)^{2} \leq \\
\leq\left(x^{2}+(\sqrt{2} y)^{2}+(\sqrt{3} z)^{2}\right) \cdot\left(1^{2}+\left(\frac{1}{\sqrt{2}}\right)^{2}+\left(\frac{1}{\sqrt{3}}\right)^{2}\right)
\end{gathered}
$$

i.e.,

$$
(x+y+z)^{2} \leq \frac{11}{6}\left(x^{2}+2 y^{2}+3 z^{2}\right)
$$

Here, the equality (which holds) is necessary and sufficient if:

$$
x: \sqrt{2} y: \sqrt{3} z=1: \frac{1}{\sqrt{2}}: \frac{1}{\sqrt{3}} .
$$

III. solution. Subtract 12 times (1) from (2) and transform it into a sum of squares:

$$
\begin{aligned}
x^{2}-12 x+2 y^{2}-12 y+3 z^{2}-12 z & =-66 \\
(x-6)^{2}+2(y-3)^{2}+3(z-2)^{2} & =0
\end{aligned}
$$

This holds if and only if $x=6, y=3, z=2$.