Conjecture
There exists infinitely many primes of the form $p_3 = p_0 + p_1 + p_2$.
Proof
Assuming these conjectures are true:
Goldbach's conjecture Every even number greater than $2$ is the sum of two primes.
$$p_0 + p_1 = 2k$$
Maillet's conjecture Every even number is the difference of two primes.
$$p_3 – p_2 = 2k$$
For $k > 2$, equating the two expressions
$$p_0 + p_1 = p_3 – p_2$$
and rearranging
$$p_3 = p_0 + p_1 + p_2$$
Since there are infinitely many $k$, there are infinitely many primes of the form $p_3 = p_0 + p_1 + p_2$ $\blacksquare$
Examples
+-----+--------+----+-----+----+--------------------------+
| k | 2k | p0 | p1 | p2 | p3 = p0 + p1 + p2 |
+-----+--------+----+-----+----+--------------------------+
| 5 | 10 | 3 | 7 | 3 | 13 = 3 + 7 + 3 |
| 50 | 100 | 3 | 97 | 7 | 107 = 3 + 97 + 7 |
| 250 | 500 | 13 | 487 | 23 | 523 = 13 + 487 + 23 |
+-----+--------+----+-----+----+--------------------------+
Best Answer
You can prove the same result on the strength of the Goldbach Conjecture alone. No need for Maillet's conjecture.
Let $p_0 \ge 7$ be any arbitrary prime number. Of course, any such number is odd, so $p_0 = 2k - 1$ for some positive integer $k \ge 4$. Now, minus the equation by $3$ to obtain
$$ p_0-3 = 2k-1-3 = 2k-4 = 2(k-2) $$
Hence, $p_0-3$ is twice some integer $k-2$, implying $p_0-3 \ge 4$ is even. Then, according to the Goldbach Conjecture, there exists two primes $p_1, p_2$ such that
$$ p_0-3 = p_1 + p_2 $$
and after adding $3$ to the equation, we have
$$ p_0 = p_1 + p_2 + 3 $$
Thus, if the Goldbach Conjecture is true, then every prime number $p_0 \ge 7$ is the sum of three primes: $3$ and the prime numbers $p_1, p_2$ such that $p_0-3 = p_1 + p_2$.