Let $ r $ be the real root. Then \[(3 - i) r^2 + (a + 4i) r - 115 + 5i = 0.\]We can write this as \[(3r^2 + ar - 115) + (-r^2 + 4r + 5)i = 0.\]The real and imaginary parts must both be 0, so $ 3r^2 + ar - 115 = 0 $ and $ -r^2 + 4r + 5 = 0 $. The equation $ -r^2 + 4r + 5 = 0 $ factors as $ -(r - 5)(r + 1) = 0, $ so $ r = 5 $ or $ r = -1 $. If $ r = 5, $ then \[3 \cdot 25 + 5a - 115 = 0.\]Solving for $ a, $ we find $ a = 8 $. If $ r = -1, $ then \[3 \cdot (-1)^2 - a - 115 = 0.\]Solving for $ a, $ we find $ a = -112 $. Thus, the possible values of $ a $ are $ \boxed{8,-112} $.