5. Given that $\boldsymbol{a}, \boldsymbol{b}$ are two mutually perpendicular unit vectors, and $|c|=13, \boldsymbol{c} \cdot \boldsymbol{a}=3, \boldsymbol{c} \cdot \boldsymbol{b}=4$. Then for any real numbers $t_{1}, t_{2}, \mid \boldsymbol{c}-t_{1} a-t_{2} b$ | the minimum value is ( ).
(A) 5
(B) 7
(C) 12
(D) 13
5.C.

From the given conditions, we have
$$
\begin{array}{l}
\left|c-t_{1} a-t_{2} b\right|^{2}=|c|^{2}-6 t_{1}-8 t_{2}+t_{1}^{2}+t_{2}^{2} \\
=169+\left(t_{1}-3\right)^{2}+\left(t_{2}-4\right)^{2}-25 \\
=144+\left(t_{1}-3\right)^{2}+\left(t_{2}-4\right)^{2} \geqslant 144 .
\end{array}
$$

When $t_{1}=3, t_{2}=4$, $\left|c-t_{1} a-t_{2} b\right|^{2}=144$.