Yes. You can easily verify this by carrying out the following steps:
First, express the means of $A$, $B$, and $C$, in terms of the model with the specified contrast:
\begin{eqnarray*}
1\hat{\beta}_{0}-2\hat{\beta}_{1}+0\hat{\beta}_{2} & = & \hat{\mu}_{A}=E(Y_{A})\\
1\hat{\beta}_{0}+1\hat{\beta}_{1}-1\hat{\beta}_{2} & = & \hat{\mu}_{B}=E(Y_{B})\\
1\hat{\beta}_{0}+1\hat{\beta}_{1}+1\hat{\beta}_{2} & = & \hat{\mu}_{C}=E(Y_{C})
\end{eqnarray*}
Here, each $\hat{\mu}_i$ represents the group mean of group $i$, $i=A, B, C$.
Next, place each beta coefficient into a matrix augmented with the
means on the right, and place the matrix in row-reduced echelon form
using Guass-Jordan elimination:
\begin{eqnarray*}
\begin{bmatrix}1 & -2 & 0 & | & \hat{\mu}_{A}\\
1 & 1 & -1 & | & \hat{\mu}_{B}\\
1 & 1 & 1 & | & \hat{\mu}_{C}
\end{bmatrix} & \sim & \begin{bmatrix}1 & -2 & 0 & | & \hat{\mu}_{A}\\
0 & 3 & -1 & | & \hat{\mu}_{B}-\hat{\mu}_{A}\\
0 & 3 & 1 & | & \hat{\mu}_{C}-\hat{\mu}_{A}
\end{bmatrix}\\
& \sim & \begin{bmatrix}1 & -2 & 0 & | & \hat{\mu}_{A}\\
0 & 3 & -1 & | & \hat{\mu}_{B}-\hat{\mu}_{A}\\
0 & 0 & 2 & | & \left(\hat{\mu}_{C}-\hat{\mu}_{A}\right)-\left(\hat{\mu}_{B}-\hat{\mu}_{A}\right)
\end{bmatrix}\\
& \sim & \begin{bmatrix}1 & -2 & 0 & | & \hat{\mu}_{A}\\
0 & 3 & -1 & | & \hat{\mu}_{B}-\hat{\mu}_{A}\\
0 & 0 & 1 & | & \frac{1}{2}\left[\left(\hat{\mu}_{C}-\hat{\mu}_{A}\right)-\left(\hat{\mu}_{B}-\hat{\mu}_{A}\right)\right]
\end{bmatrix}\\
& \sim & \begin{bmatrix}1 & -2 & 0 & | & \hat{\mu}_{A}\\
0 & 1 & 0 & | & \frac{1}{3}\left\{ \left(\hat{\mu}_{B}-\hat{\mu}_{A}\right)+\frac{1}{2}\left[\left(\hat{\mu}_{C}-\hat{\mu}_{A}\right)-\left(\hat{\mu}_{B}-\hat{\mu}_{A}\right)\right]\right\} \\
0 & 0 & 1 & | & \frac{1}{2}\left[\left(\hat{\mu}_{C}-\hat{\mu}_{A}\right)-\left(\hat{\mu}_{B}-\hat{\mu}_{A}\right)\right]
\end{bmatrix}\\
& \sim & \begin{bmatrix}1 & 0 & 0 & | & \hat{\mu}_{A}+\frac{2}{3}\left\{ \left(\hat{\mu}_{B}-\hat{\mu}_{A}\right)+\frac{1}{2}\left[\left(\hat{\mu}_{C}-\hat{\mu}_{A}\right)-\left(\hat{\mu}_{B}-\hat{\mu}_{A}\right)\right]\right\} \\
0 & 1 & 0 & | & \frac{1}{3}\left\{ \left(\hat{\mu}_{B}-\hat{\mu}_{A}\right)+\frac{1}{2}\left[\left(\hat{\mu}_{C}-\hat{\mu}_{A}\right)-\left(\hat{\mu}_{B}-\hat{\mu}_{A}\right)\right]\right\} \\
0 & 0 & 1 & | & \frac{1}{2}\left[\left(\hat{\mu}_{C}-\hat{\mu}_{A}\right)-\left(\hat{\mu}_{B}-\hat{\mu}_{A}\right)\right]
\end{bmatrix}
\end{eqnarray*}
So, now, we know that the first pivot position corresponds to:
\begin{eqnarray*}
\hat{\beta}{}_{0} & = & \hat{\mu}_{A}+\frac{2}{3}\left\{ \left(\hat{\mu}_{B}-\hat{\mu}_{A}\right)+\frac{1}{2}\left[\left(\hat{\mu}_{C}-\hat{\mu}_{A}\right)-\left(\hat{\mu}_{B}-\hat{\mu}_{A}\right)\right]\right\} \\
& = & \hat{\mu}_{A}-\frac{2}{3}\hat{\mu}_{A}-\frac{1}{3}\hat{\mu}_{A}+\frac{1}{3}\hat{\mu}_{A}+\frac{2}{3}\hat{\mu}_{B}-\frac{1}{3}\hat{\mu}_{B}+\frac{1}{3}\hat{\mu}_{C}\\
& = & \frac{1}{3}\hat{\mu}_{A}+\frac{1}{3}\hat{\mu}_{B}+\frac{1}{3}\hat{\mu}_{C}\\
& = & \frac{\hat{\mu}_{A}+\hat{\mu}_{B}+\hat{\mu}_{C}}{3}
\end{eqnarray*}
The final expression indicates that $\hat{\beta}{}_{0}$, the intercept,
represents the simple mean of the group means.