Let $ a,b,c$ be the three side-length of that triangle then $ a: b: c = \sin{\frac{\pi}{7}}: \sin{\frac{2\pi}{7}}: \sin{\frac{4\pi}{7}}$, since $ \frac{\sin{\frac{\pi}{7}}}{\sin{\frac{2\pi}{7}}}=\frac{1}{2\cos{\frac{\pi}{7}}}\not\in\mathbb{Q}$, there is no such triangle.

How to prove that $ \cos \dfrac{\pi}{7}$ is irrational? Only two contestants solve this problem by proving this, one failed, and they used rational coefficients polynomial (by expanding $ \cos \pi$ to $ \cos \dfrac{\pi}{7}$). There are many other ways to kill this problem, for example, by noting that $ a^2=b(b+c)$ and you may use this twice. Then, work in cases.

Let $ \frac {\phi}{2\pi} = \frac {m}{n},$ where $ m \ge 0, n > 0$ are integers, $ m < n$ and $ \gcd(m, n) = 1$. Assume $ 2\cos \phi = \frac {p}{q}$ is rational, where $ p$ any, $ q > 0$ are integers and $ \gcd(p, q) = 1.$ Since $ |2\cos \phi| \le 2,$ we have $ |p| \le 2q.$ Then $ 2 \cos 2 \phi = 4 \cos^{2}\phi - 2 = \frac {p^{2} - 2q^{2}}{q^{2}} = \frac {p_1}{q_1}.$ Assume that $ \gcd(p_1, q_1) = d > 1.$ Since $ d$ divides $ q_1 = q^{2},$ it divides both $ q$ and $ 2q^{2}$. Since $ d$ divides both $ p_1 = p^{2} - 2q^{2}$ and $ 2q^{2},$ it also divides $ p^{2}$ and $ p,$ so that $ \gcd(p, q) \ge d > 1,$ which is a contradiction. Therefore, $ \gcd(p_1, q_1) = 1$ as well. By induction, $ 2 \cos (2^k \phi) = \frac {p_k}{q_k},$ where $ q_k = q^{2^k}$ and $ \gcd(p_k, q_k) = 1.$ Assuming $ q > 1,$ the denominators $ q_{k} = q^{2^{k}}$ of the sequence $ \left\{\frac {p_{k}}{q_{k}}\right\} = \left\{2 \cos \left(2^{k} \cdot \frac {m}{n} \cdot 2\pi\right)\right\}$ grow without bounds for $ k \longrightarrow \infty.$ But cosine is a periodic function with period $ 2\pi$ $ \Longrightarrow$ for any $ n,$ this sequence contains at most $ n$ different values $ 2 \cos \left(\frac {l}{n}\cdot 2\pi\right),\ l = 0,\ 1,\ ...,\ n - 1,$ which contradicts $ q_{k}\longrightarrow \infty.$ As a result, $ q = 1$ and $ p = 0, \pm 1, \pm 2.$ The only angles $ \frac{\phi}{2\pi} = \frac {m}{n}$ with rational cosines are then $ \frac {\phi}{2\pi} \in \left\{0, \frac {1}{6}, \frac {1}{4}, \frac {1}{3}, \frac {1}{2}, \frac {2}{3}, \frac {3}{4}, \frac {5}{6}\right\},$ which does not include $ \frac {\phi}{2\pi} = \frac {1}{14},$ $ \phi = \frac {\pi}{7}.$

As we see, we don't need any side to be a prime number.

So it suffices to prove that $\cos \frac{\pi}{7}$ is irrationnal I will proceed by contradiction-proof assume that $\cos \frac \pi 7=x$ is rational then $\cos\frac{2\pi}{7}=2x^2-1 $ is rational and also $\cos \frac{4\pi}{7}$ is rational ; $\cos\frac{2\pi}{7}=\frac{a}{b}$ s.t. $a\wedge b=1$ thus $\cos^2\frac{\pi}{7}=\frac12(\frac{a}b+1),\cos\frac{4\pi}7=2(\frac ab)^2-1$ but we know $\cos\frac{\pi}7.\cos\frac{2\pi}7.\cos\frac{4\pi}7=-\frac1{2^3}$ thus $(\frac{\frac{a}{b}+1}{2}).(\frac ab)^2.(2(\frac{a}{b})^2-1)^2=\frac{1}{2^6}\implies 2^7. {(a+b)a^2(2a^2-b^2)^2} ={b^7}$ we know that $a$ and $b$ has different parity so $b$ is even and $a$ is odd whence $v_2$(LHS)$=9$ and $v_2$(RHS)$=7k$ which is impossible therefore the result follows . RH HAS