Ans: D
$\begin{array}{cl}
& (x+3i)(3+i) \\
= & 3x+9i+xi+3i^2 \\
= & (3x-3)+(9+x)i
\end{array}$
$\begin{array}{cl}
& (x+3i)(3+i) \\
= & 3x+9i+xi+3i^2 \\
= & (3x-3)+(9+x)i
\end{array}$
Therefore, the real part is $3x-3$.
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