If the reflection of the point $P(1,0,0)$ in the line $\frac{x-1}{2}=\frac{y+1}{-3}=\frac{z+10}{8}$ is $(\alpha ,\beta ,\gamma )$. Then  $|\alpha +\beta +\gamma |$ is

Given reflection of the point $P(1,0,0)$ in the line $\frac{x-1}{2}=\frac{y+1}{-3}=\frac{z+10}{8}$ is Q $(\alpha ,\beta ,\gamma )$
Direction ratios of  $\overline{PQ}$ is  $\alpha -1,\beta -0,\gamma -0=\alpha -1,\beta ,\gamma$
Direction ratios of the given line is  $2,-3,8$
$\overline{PQ}$  perpendicular to the given line 
(two line are perpendiculars whose direction ratios are $a_1,b_1,c_1$  and  $a_2,b_2,c_2$ then  $a_1{a}_2+b_1{b}_2+c_1{c}_2=0$)
 $$\begin{gathered} \Rightarrow \left(2\right)(\alpha -1)+(-3)\left(\beta \right)+\left(8\right)\left(\gamma \right)=0 \\ \Rightarrow 2\alpha -3\beta +8\gamma =\mathrm{2................}\left(1\right) \end{gathered}$$
Midpoint of $\overline{PQ}$ is $(\frac{\alpha +1}{2},\frac{\beta }{2},\frac{\gamma }{2})$ lies on the given line 
 $$\begin{gathered} \Rightarrow \frac{\frac{\alpha +1}{2}-1}{2}=\frac{\frac{\beta }{2}+1}{-3}=\frac{\frac{\gamma }{2}+10}{8}=\lambda \\ \Rightarrow \frac{\alpha -1}{4}=\frac{\beta +2}{-6}=\frac{\gamma +20}{16}=\lambda \\ \Rightarrow \alpha =4\lambda +1,\beta =-6\lambda -2,\gamma =16\lambda -20 \end{gathered}$$
 From (1)  
 $$\begin{gathered} \Rightarrow 2(4\lambda +1)-3(-6\lambda -2)+8(16\lambda -20)=2 \\ \Rightarrow 154\lambda =154\Rightarrow \lambda =1 \\ \therefore \alpha =5,\beta =8,\gamma =4 \\ \alpha +\beta +\gamma =7 \end{gathered}$$