958*. Solve the inequality in integers

$$
\frac{1}{\sqrt{x-2 y+z+1}}+\frac{2}{\sqrt{2 x-y+3 z-1}}+\frac{3}{\sqrt{3 y-3 x-4 z+3}}>x^{2}-4 x+3
$$
$\triangle$ The numbers $x-2y+z+1, 2x-y+3z-1$, and $3y-3x-4z+3$ are not only integers but also positive. Their sum is equal to 3.

$$
(x-2y+z+1) + (2x-y+3z-1) + (3y-3x-4z+3) = 3
$$

But when is the sum of three natural numbers equal to 3? Only in one case - when each of them is equal to 1.

$$
\left\{\begin{array}{l}
x-2y+z+1=1 \\
2x-y+3z-1=1 \\
3y-3x-4z+3=1
\end{array}\right.
$$

In the obtained system of equations, one of them (any) is a consequence of the others, since when adding all the equations of the system, we get an identity: $3=3$. Therefore, it is impossible to find the unknowns uniquely from it.

Let's use the inequality

$$
1+2+3 > x^2 - 4x + 3, \quad x^2 - 4x - 32 - 2.7 = -0.7
$$

Therefore,

$$
-0.7 < x < 4.7
$$

Then $x \in \{0; 1; 2; 3; 4\}$. Now we need to try all the obtained values of $x$, substituting them one by one into two of the three equations of the system and finding the corresponding values of $y$ and $z$. Do this on your own.

Answer: $(3; 1; -1)$.