7. Given vectors $\boldsymbol{a}, \boldsymbol{b}, \boldsymbol{c}$ satisfy
$$
\begin{array}{l}
|a|=|b|=3,|c|=4, \boldsymbol{a} \cdot \boldsymbol{b}=-\frac{7}{2}, \\
\boldsymbol{a} \perp \boldsymbol{c}, \boldsymbol{b} \perp \boldsymbol{c} .
\end{array}
$$

Then for real numbers $x, y$, the expression
$$
|x \boldsymbol{a}+y \boldsymbol{b}+(1-x-y) \boldsymbol{c}|
$$

has the minimum value of $\qquad$
7. $\frac{4 \sqrt{33}}{15}$.

From the given information,
$$
\begin{array}{l}
a^{2}=\boldsymbol{b}^{2}=9, \boldsymbol{c}^{2}=16, \\
\boldsymbol{a} \cdot \boldsymbol{b}=-\frac{7}{2}, \boldsymbol{a} \cdot \boldsymbol{c}=0, \boldsymbol{b} \cdot \boldsymbol{c}=0 .
\end{array}
$$

Then $|x \boldsymbol{a}+y \boldsymbol{b}+(1-x-y) \boldsymbol{c}|^{2}$
$$
\begin{array}{l}
=9 x^{2}+9 y^{2}+16(1-x-y)^{2}-7 x y \\
=25 x^{2}+25 y^{2}+25 x y-32(x+y)+16 \\
\geqslant 25\left(\frac{3}{4}(x+y)^{2}\right)-32(x+y)+16 \\
=\frac{75}{4}\left(x+y-\frac{64}{75}\right)^{2}+\frac{16 \times 11}{75} \\
\geqslant \frac{16 \times 11}{75},
\end{array}
$$

When $x=y=\frac{32}{75}$, the equality holds.