If the line $\frac{x-2}{1}=\frac{y-9}{2}=\frac{z-13}{3},\frac{x-a}{1}=\frac{y-1}{-2}=\frac{z+2}{3}$ are coplanar, then $a$ is 

$\overrightarrow{r}=\overrightarrow{a}+s\overrightarrow{b},\overrightarrow{r}=\overrightarrow{c}+t\overrightarrow{d}$ are coplanar if $\left[\begin{array}{llll}\overrightarrow{c}& -\overrightarrow{a}& \overrightarrow{b}& \overrightarrow{d}\end{array}\right]$

$\overrightarrow{a}=(2,9,13),\overrightarrow{c}=(a,1,-2),\overrightarrow{b}=(1,2,3),\overrightarrow{d}=(1,-2,3)$

$\therefore \left|\begin{array}{ccc}a-2& -8& -15\\ 1& 2& 3\\ 1& -2& 3\end{array}\right|\Rightarrow (a-2)12-15(-4)=0$

$\Rightarrow a-2+5=0\Rightarrow a=-3$.