7. Let the incenter of $\triangle A B C$ be $I$, and it satisfies $2 \overrightarrow{I A}+5 \overrightarrow{I B}+6 \overrightarrow{I C}=\overrightarrow{0}$, then the value of $\cos \angle B$ is $\qquad$
Connect $A I$ intersecting $B C$ at point $D$, then $\frac{B D}{D C}=\frac{c}{b}$ $\Rightarrow \overrightarrow{B D}=\frac{c}{b+c} \overrightarrow{B C}$, thus $\overrightarrow{A D}=\frac{b}{b+c} \overrightarrow{A B}+\frac{c}{b+c} \overrightarrow{A C}$.
Also, $\frac{A I}{I D}=\frac{A B}{B D}=\frac{c}{\frac{a c}{b+c}}=\frac{b+c}{a}$, therefore
$$
\overrightarrow{A I}=\frac{b+c}{a+b+c} \overrightarrow{A D}=\frac{b}{a+b+c} \overrightarrow{A B}+\frac{c}{a+b+c} \overrightarrow{A C}
$$

Similarly, we get $\overrightarrow{B I}=\frac{a}{a+b+c} \overrightarrow{B A}+\frac{c}{a+b+c} \overrightarrow{B C}$,
$\overrightarrow{C I}=\frac{a}{a+b+c} \overrightarrow{C A}+\frac{b}{a+b+c} \overrightarrow{C B}$, adding the above three equations yields $a \overrightarrow{I A}+b \overrightarrow{I B}+c \overrightarrow{I C}=\overrightarrow{0}$.
By the uniqueness of vector representation, we have $a: b: c=2: 5: 6$, so $\cos B=\frac{4+36-25}{2 \cdot 2 \cdot 6}=\frac{5}{8}$.