在等差数列 $\left\{a_{n}\right\}$ 中, $2\left(a_{1}+a_{4}+a_{7}\right)+3\left(a_{9}+a_{11}\right)=24$, 则该数列的前 13 项和 $S_{13}$ 等于
等差数列 $\left\{a_{n}\right\}$ 中, $2\left(a_{1}+a_{4}+a_{7}\right)+3\left(a_{9}+a_{11}\right)=24$,

$\therefore 6 a_{4}+6 a_{10}=24$,

$\therefore 2 a_{7}=4$, 即 $a_{7}=2$.

则此数列的前 13 项之和 $S_{13}=\frac{13\left(a_{1}+a_{13}\right)}{2}=13 a_{7}=26$.

故答案为: 26 .