Since we have at least 5 upvotes, I will provide the answer.
Hypothesis: T(n) = O(n^k), where k = log_2(13/6). That is, log to the base 2 of (13/6). We can observe that k is slightly larger than 1. It is approximately 1.115.
Now, to prove it, we simply use mathematical induction.
Hypothesis: T(n) <= c n^k
Base Case: T(1) = 1, so this is true, as long as c >= 1. (We will actually end up choosing c that is larger than that anyway.)
Induction Hypothesis: Assume T(n) <= c n^k for all n <= m
T(m) = T(m/2) + T(m/3) + T(m/4) + m
<= c m^k (1/2^k + 1/3^k + 1/4^k) + m
= c m^k 0.89 + m
We can make this to be less than equal to cm^k, as long as c > 1/0.11
So, we can choose c = 10.
(You are also allowed to do this as rough calculation before, and then start your entire hypothesis as T(n) <= 10 n^k.)