For a system of $3$ linear equations in $3$ variables to have infinitely many solutions, we need $\operatorname{rank}(A)=\operatorname{rank}([A|B])<3,$ where $A$ is the coefficient matrix and $[A|B]$ is the augmented matrix.

The system is

First, for infinitely many solutions, we must have $\det(A)=0.$

Compute the determinant:

For infinitely many solutions, the third equation must be a linear combination of the first two (since rank must be less than $3$ and consistency must hold for constants too).

Let $R_3 = pR_1 + qR_2.$

Then comparing coefficients and constants:

Solve for $p,q$ using the $x$-coefficient and constant equations: $p+2q=3$ $-4p+5q=3$

From the first, $p=3-2q.$

Substitute into the second: $-4(3-2q)+5q=3$ $-12+8q+5q=3$ $13q=15$ $q=\frac{15}{13}.$

Then $p=3-2\cdot \frac{15}{13}=\frac{39-30}{13}=\frac{9}{13}.$

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

From $-2p+\alpha q=-1,$ $-2\cdot \frac{9}{13}+\alpha\cdot \frac{15}{13}=-1.$

Multiply by $13$: $-18+15\alpha=-13$ $15\alpha=5$ $\alpha=\frac{1}{3}.$

From $\beta=p+3q,$ $\beta=\frac{9}{13}+3\cdot \frac{15}{13}=\frac{54}{13}.$

Now compute $12\alpha+13\beta = 12\left(\frac13\right)+13\left(\frac{54}{13}\right).$

$=4+54=58.$

Therefore, the correct option is $\boxed{58}.$

Compare with stored answer: Stored correct answer is D, i.e. $58$, which matches our result.