Step 1: Recall when a system has infinitely many solutions.
For three equations in three unknowns, infinitely many solutions require the coefficient determinant to be zero and the system to stay consistent.
Step 2: Build the coefficient matrix.
From $ax+y-2z=3$, $2x-y+3z=b$, $x+2y-z=3$, the matrix is $\begin{bmatrix}a&1&-2\\2&-1&3\\1&2&-1\end{bmatrix}$.
Step 3: Set its determinant to zero.
Expanding, $a(1-6)-1(-2-3)+(-2)(4+1)=0$, that is $-5a+5-10=0$.
Step 4: Solve for a.
This gives $-5a-5=0$, so $a=-1$.
Step 5: Use consistency to find b.
With $a=-1$ the three equations become dependent, and matching the right-hand sides for that dependence forces $b=-3$.
Step 6: Evaluate the asked expression.
$3a-2b=3(-1)-2(-3)=-3+6=3$, which is option (4).
\[ \boxed{3a-2b=3} \]