Part I: Find a BFS
OP has transformed the LP to this max problem.
\begin{array}{rrr}
\max z &= -3y_1 + y_2 - 2y_3 & \\
\text{s.t.} & -2y_1 + y_2 - y_3 \le& 1 \\
& -y_1 \phantom{-y_2} - 2y_3 \le& -2 \\
& 7y_1-4y_2+6y_3 \le& -1 \\
& y_1,y_2,y_3 \ge 0
\end{array}
In the given initial tableau, the problem is actually
$$
\begin{array}{rrrr}
\max z =& -3y_1 + y_2 - 2y_3 & & \\
\text{s.t.} & -2y_1 + y_2 - y_3 &+ s_1 =& 1 \\
& -y_1 \phantom{-y_2} - 2y_3 &+ s_2 =& -2 \\
& 7y_1-4y_2+6y_3 &+ s_3 =& -1 \\
& y_1,y_2,y_3,s_1,s_2,s_3 \ge 0
\end{array}
\tag{1}\label{1}
$$
The computer program handles positive variables, but the RHS shows that the current BFS is $s_1 = 1$, $s_2 = -2$ and $s_3 = -1$. This is the cause of error.
In fact, we should make RHS non-negative.
$$
\begin{array}{rrrr}
\max z =& -3y_1 + y_2 - 2y_3 & & \\
\text{s.t.} & -2y_1 + y_2 - y_3 &+ s_1 =& 1 \\
& y_1 \phantom{+y_2} + 2y_3 &- s_2 =& 2 \\
& -7y_1+4y_2-6y_3 &- s_3 =& 1 \\
& y_1,y_2,y_3,s_1,s_2,s_3 \ge 0
\end{array}
\tag{2}\label{2}
$$
We need to find feasible solution first, so we add two more artificial non-negative variables $u_1$ and $u_2$ to the LP. (To apply simplex method, we can't have negative variables.) Since $u_1$ and $u_2$ aren't in the BFS of the actual problem \eqref{2}, we eliminate them by minimizing $u_1 + u_2$.
\begin{array}{rrrr}
\max z= & -u_1 - u_2 & & \\
\text{s.t.} & -2y_1 + y_2 - y_3 &+ s_1 \phantom{+u_1} =& 1 \\
& y_1 \phantom{+y_2} + 2y_3 &- s_2 + u_1 =& 2 \\
& -7y_1+4y_2-6y_3 &- s_3 + u_2 =& 1 \\
& y_1,y_2,y_3,s_1,s_2,s_3,u_1,u_2 \ge 0
\end{array}
The actual intial simplex tableau
y_1 y_2 y_3 s_1 s_2 s_3 u_1 u_2 RHS
s_1 -2 1 -1 1 0 0 0 0 1
u_1 1 0 2 0 -1 0 1 0 2
u_2 -7 4 -6 0 0 -1 0 1 1
---------------------------------------
0 0 0 0 0 0 1 1 0
y_1 y_2 y_3 s_1 s_2 s_3 u_1 u_2 RHS
s_1 -2 1 -1 1 0 0 0 0 1
u_1 1 0 2 0 -1 0 1 0 2
u_2 -7 4* -6 0 0 -1 0 1 1
---------------------------------------
6 -4 4 0 1 1 0 0 -3
y_1 y_2 y_3 s_1 s_2 s_3 u_1 u_2 RHS
s_1 -1/4 0 1/2 1 0 1/4 0 -1/4 3/4
u_1 1 0 2* 0 -1 0 1 0 2
y_2 -7/4 1 -3/2 0 0 -1/4 0 1/4 1/4
--------------------------------------------
-1 0 -2 0 1 0 0 1 -2
y_1 y_2 y_3 s_1 s_2 s_3 u_1 u_2 RHS
s_1 -1/2 0 0 1 1/4 1/4 -1/4 -1/4 1/4
y_3 1/2 0 1 0 -1/2 0 1/2 0 1
y_2 -1 1 0 0 -3/4 -1/4 3/4 1/4 7/4
--------------------------------------------
0 0 0 0 0 0 1 1 0
Now we've a BFS $y_2 = \frac74$, $y_3 = 1$ and $s_1 = \frac14$.
Part II: Find an optimal BFS
Return to \eqref{2}.
y_1 y_2 y_3 s_1 s_2 s_3 RHS
s_1 -1/2 0 0 1 1/4 1/4 1/4
y_3 1/2 0 1 0 -1/2 0 1
y_2 -1 1 0 0 -3/4 -1/4 7/4
----------------------------------
3 -1 2 0 0 0 0
y_1 y_2 y_3 s_1 s_2 s_3 RHS
s_1 -1/2 0 0 1 1/4 1/4* 1/4
y_3 1/2 0 1 0 -1/2 0 1
y_2 -1 1 0 0 -3/4 -1/4 7/4
-----------------------------------
1 0 0 0 1/4 -1/4 -1/4
y_1 y_2 y_3 s_1 s_2 s_3 RHS
s_3 -2 0 0 4 1 1 1
y_3 1/2 0 1 0 -1/2 0 1
y_2 -3/2 1 0 1 -1/2 0 2
---------------------------------
1/2 0 0 1 1/2 0 0
Hence, the optimal solution to \eqref{2} (thus the original LP) is $(y_1,y_2,y_3,s_1,s_2,s_3) = (0,2,1,0,0,1)$ with optimal value 0.
Best Answer
No, this need not be true. In case of degeneracy, you will change the basis but not improve the objective value. Geometrically this means that you are in a vertex of the polyhedron describing the feasible set and the change of basis does not lead you to a new vertex. Hence, you are stuck in the same vertex for one (or possibly more) iterations and you do not improve the objective value. Degeneracy occurs when there are redundant constraints in the problem.