Given the system of linear equations.

$8x+y+4z=-2$

$x+y+z=0$

$\lambda x-3y=\mu$

has infinitely many solutions, so

$D=\left|\begin{array}{ccc}8& 1& 4\\ 1& 1& 1\\ \lambda & -3& 0\end{array}\right|=0\Rightarrow \lambda =4$

Also $D_1=D_2=D_3=0$

$D_1=\left|\begin{array}{ccc}-2& 1& 4\\ 0& 1& 1\\ \mu & -3& 0\end{array}\right|=0\Rightarrow -6-3\mu =0$

So $\mu =-2$

Hence the point will be $\left(4,-2,-\frac{1}{2}\right)$

Now distance of point $\left(4,-2,-\frac{1}{2}\right)$ from plane $8x+y+4z+2=0$

$=\left|\frac{32-2-2+2}{\sqrt{8^2+1^2+4^2}}\right|=\left|\frac{30}{9}\right|=\frac{10}{3}$