= Solution
Choose <Cramer's rule>, the permitted alternative to finding an inverse. The <determinant> is
$$
\Delta=1(2-1)-2(-1-2)+3(1+4)=22\ne0.
$$
Replacing each column by the right-hand side gives
$$
\Delta_x=\det\begin{pmatrix}14&2&3\\0&-2&1\\1&1&-1\end{pmatrix}
=14+2+6=22,
$$
$$
\Delta_y=\det\begin{pmatrix}1&14&3\\1&0&1\\2&1&-1\end{pmatrix}
=-1+42+3=44,
$$
$$
\Delta_z=\det\begin{pmatrix}1&2&14\\1&-2&0\\2&1&1\end{pmatrix}
=-2-2+70=66.
$$
Consequently
$$
\boxed{x=\Delta_x/\Delta=1,\quad y=\Delta_y/\Delta=2,\quad z=\Delta_z/\Delta=3.}
$$
The nonzero <determinant> also proves uniqueness.
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