For the system to have infinitely many solutions, we need:

• The coefficient matrix to be singular: $\det(A)=0$
• And the system to be consistent, i.e. ranks of coefficient and augmented matrices must be equal and less than 3.

The system is

Since infinitely many solutions require one equation to be a linear combination of the other two, let us use the third equation to express such dependence.

Suppose $R_1=aR_3+bR_2$ in the augmented sense.

Comparing coefficients and constants: $3=a+2b$ $1=2a+\alpha b$ $\beta=a-b$ $3=4a-3b$

Solve for $a,b$ using the first and fourth equations: $a+2b=3$ $4a-3b=3$

From the first, $a=3-2b$

Substitute into the second: $4(3-2b)-3b=3$ $12-8b-3b=3$ $12-11b=3$ $11b=9$ $b=\frac{9}{11}$

Hence $a=3-2\cdot\frac{9}{11}=\frac{33-18}{11}=\frac{15}{11}.$

Now find $\alpha$ and $\beta$.

From $1=2a+\alpha b$ we get $1=2\cdot\frac{15}{11}+\alpha\cdot\frac{9}{11}$ $1=\frac{30}{11}+\frac{9\alpha}{11}$ $11=30+9\alpha$ $9\alpha=-19$ $\alpha=-\frac{19}{9}.$

From $\beta=a-b$ $\beta=\frac{15}{11}-\frac{9}{11}=\frac{6}{11}.$

Compute the required value: $22\beta-9\alpha=22\cdot\frac{6}{11}-9\left(-\frac{19}{9}\right)$ $=12+19=31.$

Therefore, the correct option is $\boxed{31}$ which is Option A.

Comparison with stored answer:

Stored correct answer = A

Our derived answer = A

So they agree.