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

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

$\text{ Direction ratios of }\overline{PQ}\text{ is }\alpha -1,\beta -0,\gamma -0=\alpha -1,\beta ,\gamma$

Direction ratios of the given line is $2,-3,8$

 Perpendicular to the given line $\overline{PQ}$

$\Rightarrow \left(2\right)\left(\alpha -1\right)+\left(-3\right)\left(\beta \right)+\left(8\right)\left(\gamma \right)=0$

$\text{ ( }\because \text{ Two lines are perpendicular whose direction ratios are }{a}_1,b_1,c_1\text{ and }{a}_2,b_2,c_2\text{ then }$$a_1{a}_2+b_1{b}_2+c_1{c}_2=0)$

$\Rightarrow 2\alpha -3\beta +8\gamma =2\dots \dots \dots \dots \dots \left(1\right)$

$\text{ Midpoint of }\overline{PQ}\text{ is }\left(\frac{\alpha +1}{2},\frac{\beta }{2},\frac{\gamma }{2}\right)\text{ lies on the given line }\Rightarrow \frac{\frac{\alpha +1}{2}-1}{2}=\frac{\frac{\beta }{2}+1}{-3}=\frac{\frac{\gamma }{2}+10}{8}=\lambda$

$\begin{array}{l}\Rightarrow \frac{\alpha -1}{4}=\frac{\beta +2}{-6}=\frac{\gamma +20}{16}=\lambda \\ \Rightarrow \alpha =4\lambda +1,\beta =-6\lambda -2,\gamma =16\lambda -20\\ \text{ From }\left(1\right)\Rightarrow 2(4\lambda +1)-3(-6\lambda -2)+8(16\lambda -20)=2\\ \Rightarrow 154\lambda =154\\ \Rightarrow \lambda =1\\ \therefore \alpha =5,\beta =-8,\gamma =-4\\ \alpha +\beta +\gamma =-7\end{array}$