3. Depending on the real parameter $k$, determine how many solutions $(x, y, z)$ in the set of real numbers the equation has

$$
\sqrt{x^{2}-6 x z+6 x+9 z^{2}-18 z+9}+|2 x+k y-z+2|+(x+2 y+k z-1)^{20190316}=0
$$
3. The given expression can be transformed as follows: $x^{2}-6 x z+6 x+9 z^{2}-18 z+9=(x-3 z)^{2}+6(x-3 z)+9=(x-3 z+3)^{2}$, so the first term is actually $|x-3 z+3|$. Note that all the terms are non-negative, so for their sum to be 0, all must be equal to zero. Therefore, the equation in the problem is actually equivalent to the following system:

$$
\begin{aligned}
x \quad-3 z & =-3 \\
2 x+k y-z & =-2 \\
x+2 y+k z & =1
\end{aligned}
$$

We multiply the first equation by $-\frac{1}{3}$ and add it to the second, and multiply it by $\frac{k}{3}$ and add it to the third, reducing them to $\frac{5}{3} x+k y=-1$ and $\frac{k+3}{3} x+2 y=1-k$. We then multiply the first of these two equations by $-\frac{k+3}{5}$ and add it to the next, reducing it to $\frac{10-k^{2}-3 k}{5} y=\frac{8-4 k}{5}$, or $(10-k^{2}-3 k) y=8-4 k$. Therefore, the system is equivalent to

$$
\begin{aligned}
x-3 z & =-3 ; \\
\frac{5}{3} x+k y & =-1 ; \\
\left(10-k^{2}-3 k\right) y & =8-4 k
\end{aligned}
$$

For $10-k^{2}-3 k \neq 0$ the system has a unique solution (from the last equation we calculate $y$, from the previous one $x$, and then from the first $z$). The expression on the left side is 0 for $k=\frac{3 \pm \sqrt{(-3)^{2}+4 \cdot 10}}{-2}=\frac{3 \pm 7}{-2}$, i.e., for $k=-5$ and $k=2$. Therefore, for $k \neq-5,2$ the system has a unique solution. For $k=-5$ the last equation reduces to $0=28$, so the system has no solution, and for $k=2$ the last equation is satisfied for any $y$ (and for each $y$ we can find corresponding $x$ and $z$), so the system has infinitely many solutions.