Step 1: Approach
Place the repeated letters one kind at a time using combinations.
Step 2: Choose positions
From 11 places choose 2 for I: $\binom{11}{2}=55$. From the remaining 9 choose 3 for N: $\binom{9}{3}=84$. From the remaining 6 choose 3 for T: $\binom{6}{3}=20$.
Step 3: Remaining letters
The 3 spaces left are filled by S, U, O in $3!=6$ ways.
Step 4: Total
\[ 55\times84\times20\times6=554400 \]
This is option (D).
Final Answer:
The word has 11 letters with I twice, N three times and T three times, giving 554400 arrangements, option (D).
\[ \boxed{554400} \]