943. Solve in integers $x, y$ and $z$ the inequality

$$
x^{2}+y^{2}+z^{2}+3 < x y + 3 y + 2 z
$$
$\triangle$ Let's gather all the terms of the inequality on the left side:

$$
x^{2}+y^{2}+z^{2}-x y-3 y-2 z+3<0
$$

Replace this inequality with the following:

$$
x^{2}+y^{2}+z^{2}-x y-3 y-2 z+4 \leqslant 0
$$

The obtained inequality is equivalent to the previous one for integer $x, y,$ and $z$ (think about why).

Now, let's isolate the squares of sums on the left side. It is useful to multiply the inequality by 4 first.

$$
\begin{gathered}
4 x^{2}+4 y^{2}+4 z^{2}-4 x y-12 y-8 z+16 \leqslant 0 \\
\left(4 x^{2}-4 x y+y^{2}\right)+\left(4 z^{2}-8 z+4\right)+\left(3 y^{2}-12 y+12\right) \leqslant 0 \\
(2 x-y)^{2}+(2 z-2)^{2}+3(y-2)^{2} \leqslant 0
\end{gathered}
$$

It is clear that the sum on the left side of the last inequality cannot be negative; it can only equal zero. Therefore,

$$
\left\{\begin{array}{l}
2 x-y=0 \\
2 z-2=0 \\
y-2=0
\end{array}\right.
$$

From here, we find $x, y,$ and $z$:

$$
y=2, \quad x=1, \quad z=1
$$

Answer: $(1 ; 2 ; 1)$.