What you want is called a weighted average, and you have calculated the important parts already, what you call "a_percent" and "b_percent". The formula you are looking for is
$$\text{average}_\text{weighted} =\text{a_percent}\times a_1 + \text{b_percent}\times b_1$$
$$\text{average}_\text{weighted} =0.567\times 0.35 + 0.433\times 0.52 \approx 0.42.$$
Since the relative weights ($0.567$ and $0.433$) are near $0.5$, the weighted average is near the 'normal' average ($\frac{a_1+b_1}2 = 0.5\times a_1 + 0.5 \times b_1 = 0.435$).
Since the weight for $a_1$ is bigger than $0.5$, one would presume that the weigthed average is slightly nearer to $a_1$ than the normal average, which is correct.
If you multiply the first ratio $(1:2:3)$ by $a$, the second by $b$, and the third by $c$, then sum these ratios together, you get $a+3b+2c : 2a + 7b + c : 3a + b + 2 c$, and you are finding $a,b,c$ such that this ratio is $3 : 4 : 5$.
Therefore you have three equations in three unknowns below:
$$a + 3b + 2c = 3$$
$$2a + 7b + c = 4$$
$$3a + b + 2c = 5$$
where $a,b,c$ are not necessarily integers.
You can use matrices to solve for this system, or you can just isolate one of the variables, bringing it down to two equations in two unknowns if you want to do this by hand.
Alternatively, a GDC (graphics display calculator) or WolframAlpha will solve the system of equations for you, giving $(a, b, c) = (\frac{8}{7}, \frac{1}{7}, \frac{5}{7})$. Since the common denominator is $7$, multiplying $3:4:5$ by $7$ gives $21:28:35$.
Best Answer
Just like chained equalities, you can break it apart. $x:3:y=-2:3:-4$ becomes $x:3=-2:3$ and $3:y=3:-4$. Now solve them independently.