7.4. Determine the natural prime numbers $a, b, c$, knowing that:

$$
\frac{2 a+4 b-c+7}{a+3 b+c+2}=\frac{5 a+b+3 c-22}{a-2 b+2 c+30}=\frac{c-3 b-6 a-14}{3 b+3 c-90}
$$
## Solution:

According to the property of equal ratio sequences, by adding the numerators and denominators to each other, we obtain:

$$
\frac{2 a+4 b-c+7}{a+3 b+c+2}=\frac{5 a+b+3 c-22}{a-2 b+2 c+30}=\frac{c-3 b-6 a-14}{3 b+3 c-90}=\frac{a+2 b+3 c-29}{2 a+4 b+6 c-58}=\frac{1}{2}
$$

Thus, we get

$$
\frac{2 a+4 b-c+7}{a+3 b+c+2}=\frac{5 a+b+3 c-22}{a-2 b+2 c+30}=\frac{c-3 b-6 a-14}{3 b+3 c-90}=\frac{1}{2}
$$

From $\frac{2 a+4 b-c+7}{a+3 b+c+2}=\frac{1}{2} \Rightarrow 4 a+8 b-2 c+14=a+3 b+c+2 \Rightarrow 3 a+5 b-3 c+12=0$. From the last equation, we deduce $b: 3$. Since $b$ is prime, it follows that $b=3$. For $b=3$, we get $3 a-3 c+27=0 \Rightarrow 3 c-3 a=27 \Rightarrow c-a=9$. From the last equality, it follows that $a$ and $c$ have different parities. Thus, $a=2, c=11$.

Answer: $a=2, b=3, c=11$