Example 3 Given that $\boldsymbol{a}$ and $\boldsymbol{b}$ are two mutually perpendicular unit vectors, and $|\boldsymbol{c}|=13, \boldsymbol{c} \cdot \boldsymbol{a}=3, \boldsymbol{c} \cdot \boldsymbol{b}=4$, then for any real numbers $t_{1}, t_{2},\left|\boldsymbol{c}-t_{1} \boldsymbol{a}-t_{2} \boldsymbol{b}\right|$ the minimum value is
A. 5
B. 7
C. 12
D. 13
Analyze According to the problem, we get $|\boldsymbol{a}|=|\boldsymbol{b}|=1$, and $\boldsymbol{a} \cdot \boldsymbol{b}=0$, thus $\left|\boldsymbol{c}-t_{1} \boldsymbol{a}-t_{2} \boldsymbol{b}\right|^{2}=c^{2}+t_{1}^{2} a^{2}+$ $t_{2}^{2} \boldsymbol{b}^{2}-2 t_{1} \boldsymbol{c} \cdot \boldsymbol{a}-2 t_{2} \boldsymbol{c} \cdot \boldsymbol{b}+2 t_{1} t_{2} \boldsymbol{a} \cdot \boldsymbol{b}=t_{1}^{2}+t_{2}^{2}-6 t_{1}-8 t_{2}+169=\left(t_{1}-3\right)^{2}+\left(t_{2}-4\right)^{2}+144 \geqslant 144$, so $\left|\boldsymbol{c}-t_{1} \boldsymbol{a}-t_{2} \boldsymbol{b}\right| \geqslant 12$, when and only when $t_{1}=3, t_{2}=4$, the above equation takes the equal sign, hence the minimum value sought is 12. Therefore, the answer is $\mathrm{C}$.