We have,
$A B C$ be a triangle with vertices
$\begin{aligned}
& A(\alpha, 5, \beta), B(-2,1,6) \text { and } C(1,0,-3) \\
& D=\text { Mid-point of } A C \\
& \therefore D\left(\frac{\alpha+1}{2}, \frac{5}{2}, \frac{\beta-3}{2}\right)
\end{aligned}$

$\begin{aligned}
& \mathbf{B D}=\left(\frac{\alpha+1}{2}+2\right) \hat{\mathbf{i}}+\left(\frac{5}{2}-1\right) \hat{\mathbf{j}}+\left(\frac{\beta-3}{2}-6\right) \hat{\mathbf{k}} \\
& \mathbf{B D}=\left(\frac{\alpha+5}{2}\right) \hat{\mathbf{i}}+\left(\frac{3}{2}\right) \hat{\mathbf{j}}+\left(\frac{\beta-15}{2}\right) \hat{\mathbf{k}}
\end{aligned}$
BD is equally inclined to the coordinate axis
$\begin{aligned}
& \therefore\left(\frac{\alpha+5}{2}\right)^2=\left(\frac{3}{2}\right)^2=\left(\frac{\beta-15}{2}\right)^2 \\
& \Rightarrow \frac{\alpha+5}{2}=\frac{3}{2} \text { and } \frac{\beta-15}{2}=\frac{3}{2} \\
& \Rightarrow \alpha=-2 \text { and } \beta=18 \\
& \text {and } \alpha+\beta=-2+18=16
\end{aligned}$