For the system to have infinitely many solutions, the three equations must be dependent and consistent.

So, the coefficient matrix must have determinant zero, and the third equation should be a linear combination of the first two.

The equations are:

Let the third equation be a linear combination of the first two: $\text{Eq.3}=p(\text{Eq.1})+q(\text{Eq.2})$

Comparing coefficients of $x,y,z$ and constants:

From $x$: $p+2q=-1$

From $y$: $p+3q=-3$

Subtracting, $q=-2$ Then $p+2(-2)=-1 \Rightarrow p=3$

Now compare the $z$-coefficients: $2p+aq=b$ $2(3)+a(-2)=b$ $6-2a=b$

Compare constants: $6p+(a+1)q=2b$ $6(3)+(a+1)(-2)=2b$ $18-2a-2=2b$ $16-2a=2b$ $8-a=b$

Use both expressions for $b$: $6-2a=8-a$ $-a=2$ $a=-2$

Then $b=8-(-2)=10$

Now compute: $7a+3b=7(-2)+3(10)=-14+30=16$

Therefore, the correct option is: $\boxed{16}$ i.e. Option D.