It's like a Human Problem: too many Unknowns and too few Solutions

(It's on the same page as yesterday's post.)
      In translation :

                          "Determine all triplets $(x,y,z)$ of integers that satisfy the inequality

$$3x^2+y^2+z^2-3xy-y-2z+1<0."$$

ANSWER CiP

Only one solution    $x=1,\;\;y=2,\;\;z=1$

                           Solution CiP

          We multiply the inequality by 12 and group the terms, obtaining the equivalent

$36x^2+\underline{12y^2}+12z^2-\underline{\underline{36xy}}-12y-24z+12<0\;\;\Leftrightarrow$

$\Leftrightarrow\;9(4x^2-\underline{\underline{4xy}}+\underline{y^2})+\underline{3y^2-12y}+12z^2-24z+\underline{12}<0\;\;\Leftrightarrow$

$\Leftrightarrow\;9(2x-y)^2+3(y^2-4y+4)+12z^2-24z<0\;\;\;|+12$

$\Leftrightarrow\;9(2x-y)^2+3(y-2)^2+12(z-1)^2<12 \tag{1}$

     If $z-1 \neq 0$, then because $9(2x-y)^2+3(y-2)^2 \geqslant 0$, (1) is not satisfied.

     So $z=1$ and (1) becomes

$3(2x-y)^2+(y-2)^2<4 \tag{2}$

     If $y=2$ then from (2) result $2x-y=0\;or\;\pm 1$. In the first case $2x-2=0$ so $x=1$; in the second case $y=2x\pm1=2$, impossible  with integers.

      If $(y-2)^2 \geqslant 1$  then $3(2x-y)^2<4-(y-2)^2\leqslant 3$ so $2x-y=0\;or\;\pm1$. In the first case $y=2x$ and (2) becomes $(y-2)^2<4$ so $y=2$ and hence $x=1$, or $y\in \{1,\;3\}$ wich cannot be equal to $2x$. In the second case $(y-2)^2\underset{(2)}{<}4-3$ so $y=2$, but $2x-2=\pm1$ is impossible with integers.

$\blacksquare$