Consider the set $T$ of all triangles whose sides are distinct prime numbers which are also in arithmetic progression. Let $\Delta \in T$ be the triangle with the least perimeter. If $a$ is the largest angle of $\Delta$ (in degrees) and $L$ is its perimeter, determine the value of $a/L$.

Step 1: Let the sides of the triangle be prime numbers $p_1<p_2<p_3$ in arithmetic progression. By the triangle inequality, we must have $p_1+p_2>p_3$.

To find the triangle with the least perimeter, we test small prime progressions:

• If $p_1=2$: the progression must be $(2,2+d,2+2d)$. For all three to be primes, $d$ must be even. But if $d$ is even, $2+d$ is even, so not a prime (since $2+d>2$). Thus, $p_1$ cannot be 2.

• If $p_1=3$: let's try progression with common difference $d=2$: $(3,5,7)$. Check triangle inequality: $3+5=8>7$, which is valid! The perimeter is $L=3+5+7=15$.

Any other prime progression (e.g., common difference $d\ge 4$ or starting with larger primes) will have a larger perimeter.
So $\Delta$ has sides $3,5,7$ and perimeter $L=15$.

Step 2: Find the largest angle $a$ of the triangle. The largest angle is opposite the longest side $7$.
By the Law of Cosines:

Since $a\in (0,180^{\circ })$, we have $a=120^{\circ }$.

Step 3: Calculate $a/L$:

Therefore, the value of $a/L$ is $8$.