Cramer's rule for a 3×3 system

Determinant of $A$.
$$\det A = \begin{vmatrix} 1 & 1 & 1 \\ 2 & -1 & 1 \\ 1 & 2 & -1 \end{vmatrix} = 1(1-2)-1(-2-1)+1(4+1)=-1+3+5=7.$$

Determinant $A_1$ (replace column 1 by $B=(6,3,2)^T$):
$$\det A_1 = \begin{vmatrix} 6 & 1 & 1 \\ 3 & -1 & 1 \\ 2 & 2 & -1 \end{vmatrix} = 6(1-2)-1(-3-2)+1(6+2)=-6+5+8=7.$$

Determinant $A_2$ (replace column 2 by $B$):
$$\det A_2 = \begin{vmatrix} 1 & 6 & 1 \\ 2 & 3 & 1 \\ 1 & 2 & -1 \end{vmatrix} = 1(-3-2)-6(-2-1)+1(4-3)=-5+18+1=14.$$

Determinant $A_3$ (replace column 3 by $B$):
$$\det A_3 = \begin{vmatrix} 1 & 1 & 6 \\ 2 & -1 & 3 \\ 1 & 2 & 2 \end{vmatrix} = 1(-2-6)-1(4-3)+6(4+1)=-8-1+30=21.$$

Solution.
$$x=\frac77=1, \qquad y=\frac{14}{7}=2, \qquad z=\frac{21}{7}=3.$$

Verification. $1+2+3=6$, $2-2+3=3$, $1+4-3=2$. Correct.