Given, $A=\left[\begin{array}{cccc}1 & 2 & 3 & 0 \\ 2 & 4 & 3 & 2 \\ 3 & 2 & 1 & 3 \\ 6 & 8 & 7 & \alpha\end{array}\right]$
Applying $R_2 \rightarrow R_2-2 R_1, \quad R_3 \rightarrow R_3-3 R_1 \quad$ and
$$
\begin{aligned}
R_4 \rightarrow R_4 & -2 R_3 \\
= & {\left[\begin{array}{rrrr}
1 & 2 & 3 & 0 \\
0 & 0 & -3 & 2 \\
0 & -4 & -8 & 3 \\
0 & 4 & 5 & \alpha-6
\end{array}\right] }
\end{aligned}
$$
Applying $R_4 \rightarrow R_4+R_3$
$$
=\left[\begin{array}{rrrr}
1 & 2 & 3 & 0 \\
0 & 0 & -3 & 2 \\
0 & -4 & -8 & 3 \\
0 & 0 & -3 & \alpha-3
\end{array}\right]
$$
Applying $R_4 \rightarrow R_4-R_2$
$$
=\left[\begin{array}{rrrr}
1 & 2 & 3 & 0 \\
0 & 0 & -3 & 2 \\
0 & -4 & -8 & 3 \\
0 & 0 & 0 & \alpha-5
\end{array}\right]
$$
Since, the matrix $A$ is of rank 3 .
$$
\therefore \quad \alpha-5=0 \Rightarrow \alpha=5
$$