= Solution
{c}
Perform the <elementary row operations> $R_2\leftarrow R_2+2R_1$ and $R_3\leftarrow R_3-R_1$, then $R_2\leftarrow R_2/2$, $R_1\leftarrow R_1-R_2$, and finally $R_1\leftarrow R_1+R_3$, $R_2\leftarrow R_2-3R_3$:
$$
\begin{pmatrix}1&1&2\\-2&0&2\\1&1&3\end{pmatrix}
\longrightarrow
\begin{pmatrix}1&1&2\\0&2&6\\0&0&1\end{pmatrix}
\longrightarrow
\begin{pmatrix}1&0&-1\\0&1&3\\0&0&1\end{pmatrix}
\longrightarrow I_3.
$$
Consequently the <reduced row echelon form> is $I_3$. Applying the same operations to the augmented column $b=(b_1,b_2,b_3)^T$ gives
$$
\boxed{v=\begin{pmatrix}
b_3-b_1-\tfrac12b_2\\
4b_1+\tfrac12b_2-3b_3\\
b_3-b_1
\end{pmatrix}.}
$$
Thus \b[every] $b\in\mathbb R^3$ admits exactly one solution. This follows because the <elementary row operations> preserve the solution set of the augmented <linear system>, and the reduced coefficient <matrix> is the <identity matrix>.
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