We need to consider several possibilities.

If $x < -a$, then $-2(x - a) - 3(x + a) = 1$, which implies $x = -\frac{a+1}{5}$. For this to be a valid solution, we must have

$-\frac{a+1}{5} < -a$, or $a < \frac{1}{4}$.

If $-a \le x \le a$, then $-2(x-a) + 3(x+a) = 1$, so $x = 1-5a$. This solution is valid only when $-a \le 1-5a \le a$, which is equivalent to $\frac{1}{6} \le a \le \frac{1}{4}$.

If $x > a$, then $2(x-a) + 3(x+a) = 1$, or $x = \frac{1-a}{5}$. For this solution to be valid we must have $\frac{1-a}{5} > a$, or $a < \frac{1}{6}$.

The equation has at least one real solution if and only if $a \le \frac{1}{4}$.