Yes, in order to put this model into standard form, we must add slack and artificial variables. For simplicity, we'll make the righthand-side non-negative.
Thus, given the model,
$$\max\quad z=-2x_1-x_2-x_3$$
Subject to,
$$x_1+x_3\ge1$$
$$x_1-x_2\ge2$$
$$x_2-x_3\ge1$$
$$x_1,x_2,x_3\ge0$$
Put in it into standardized form (Big-M) as such:
$$\max\quad z+2x_1+x_2+x_3+M(a_1+a_2+a_3)=0$$
Subject to,
$$x_1+x_3-e_1+a_1=1$$
$$x_1-x_2-e_2+a_2=2$$
$$x_2-x_3-e_3+a_3=1$$
$$x_1,x_2,x_3,e_1,e_2,e_3,a_1,a_2,a_3\ge0$$
Where $M$ is an arbitrary "largest" number in $\Bbb R$.
This standard form will look like the following in the initial tableau:
\begin{array} {|c|c|}
\hline BV & z & x_1 & x_2 & x_3 & e_1 & e_2 & e_3 & a_1 & a_2 & a_3 & RHS & RT \\
\hline z & 1 & 2 & 1 & 1 & 0 & 0 & 0 & M & M & M & 0 & - \\
\hline ? & 0 & 1 & 0 & 1 & -1 & 0 & 0 & 1 & 0 & 0 & 1 & - \\
? & 0 & 1 & -1 & 0 & 0 & -1 & 0 & 0 & 1 & 0 & 2 & - \\
? & 0 & 0 & 1 & -1 & 0 & 0 & -1 & 0 & 0 & 1 & 1 & - \\ \hline
\end{array}
Since there are no basic variables in this tableau, we'll need to make $a_1,a_2,$ and $a_3$ our initial basic variables as shown in the following tableau:
\begin{array} {|c|c|}
\hline BV & z & x_1 & x_2 & x_3 & e_1 & e_2 & e_3 & a_1 & a_2 & a_3 & RHS & RT \\
\hline z & 1 & -2M+2 & 1 & 1 & M & M & M & 0 & 0 & 0 & -4M & - \\
\hline a_1 & 0 & 1 & 0 & 1 & -1 & 0 & 0 & 1 & 0 & 0 & 1 & 1 \\
a_2 & 0 & 1 & -1 & 0 & 0 & -1 & 0 & 0 & 1 & 0 & 2 & 1 \\
a_3 & 0 & 0 & 1 & -1 & 0 & 0 & -1 & 0 & 0 & 1 & 1 & \infty \\ \hline
\end{array}
From here, the $x_1$ column has the most negative $C^\pi_j$ for this maximization problem, thus we'll pivot and carry on like any other model via the Simplex method.