812. Solve in integers $x, y$ and $z$ the equation

$$
5 x^{2} + y^{2} + 3 z^{2} - 2 y z = 30
$$
$\triangle$ Transform the equation by completing the square for the difference $y^{2}-2 y z$:

$$
5 x^{2}+\left(y^{2}-2 y z+z^{2}\right)+2 z^{2}=30, \quad 5 x^{2}+(y-z)^{2}+2 z^{2}=30
$$

From this, it is clear that the unknowns can only take values with small absolute values. Let's determine the constraints for $x$:

$$
5 x^{2} \leqslant 30, \quad x^{2} \leqslant 6, \quad|x| \leqslant 2, \quad-2 \leqslant x \leqslant 2
$$

For now, let's consider non-negative values of $x$. Then $0 \leqslant x \leqslant 2$. We will consider three cases.

1) Let $x=0$. The equation becomes:

$$
(y-z)^{2}+2 z^{2}=30
$$

Find the constraints for $z$:

$$
2 z^{2} \leqslant 30, \quad z^{2} \leqslant 15, \quad|z| \leqslant 3
$$

If $z^{2}=0,1,4,9$, then from the last equation, respectively:

$$
(y-z)^{2}=30,28,22,18
$$

In none of these four cases are there integer solutions.

2) Let $x=1$. We get:

$$
5+(y-z)^{2}+2 z^{2}=30, \quad(y-z)^{2}+2 z^{2}=25
$$

Similarly, we establish that only $z=0, y= \pm 5$ fits.

3) Let $x=2$. Then:

$$
20+(y-z)^{2}+2 z^{2}=30, \quad(y-z)^{2}+2 z^{2}=10
$$

Similarly, we determine that the last equation has no integer solutions.

When writing the answer, note that $x$ can also be negative. For $x=-1$, we get the same values of $y$ and $z$ as for $x=1$.

Answer: $(1 ; 5 ; 0),(1 ;-5 ; 0),(-1 ; 5 ; 0),(-1 ;-5 ; 0)$