The reflection of the point $\text{A}(1, 0, 0)$ in the line $\frac{x-1}{2}=\frac{y+1}{-3}=\frac{z+10}{8}$ is 

Any point P on the given line is  $(2r+1,-3r-1,8r-10)$
So the direction ratios of AP are $2r, -3r-1, 8r – 10$
Now AP is perpendicular to the given line if 
$$\begin{gathered} 2\left(2r\right)-3(-3r-2)+8(8r-10)=0 \\ \Rightarrow 77r-77=0\Rightarrow r=1 \end{gathered}$$ 
And thus the coordinates of P, the foot of the perpendicular from A on the line are $(3, -4, -2)$
Let $B(a, b, c)$be the reflection of A in the given line. Then P is the mid-point of AB
$\frac{a+1}{2}=3,\frac{b}{2}=-4,\frac{c}{2}=-2\Rightarrow a=5,b=-8,c=-4$ 
Thus, the coordinates of required point are $(5, -8, -4)$