Consider the line $L:\frac{x-1}{2}=\frac{y+1}{-3}=\frac{z+10}{8}$ and a family of planes $P$ containing the line $L$. The member of the family of planes $P$ which is situated at a maximum distance from $A\left(1,0,0\right)$ will be

Let the foot of the perpendicular from $A\left(1,0,0\right)$ to given line $L=0$ is $B$, then the required plane must be perpendicular to $AB$ and contain the given line.
Any general point on the given line is $B=\left(2\lambda +1,-3\lambda -1,8\lambda -10\right)$
$\overrightarrow{AB}=<2\lambda ,-3\lambda -1,8\lambda -10>$
$\therefore \overrightarrow{AB}$ is perpendicular to the given line
$\Rightarrow 2\left(2\lambda \right)+\left(-3\right)\left(-3\lambda -1\right)+8\left(8\lambda -10\right)=0$
$\Rightarrow 4\lambda +9\lambda +3+64\lambda -80=0$
$\Rightarrow \lambda =1$
So, the coordinates of point $B$ are $\left(3,-4,-2\right)$
Equation of the required plane is
$2\left(x-1\right)-4\left(y+1\right)-2\left(z+10\right)=0$
$\Rightarrow 2x-2-4y-4-2z-20=0$
$\Rightarrow 2x-4y-2z-26=0$
$\Rightarrow x-2y-z=13$