If the system of linear equations $8x+y+4z=-2,$ $x+y+z=0,$ $\lambda x-3y=\mu$ has infinitely many solutions then the distance of the point $\left(\lambda ,\mu ,\frac{-1}{2}\right)$ from the plane $8x+y+4z+2=0$ is

Given equation are

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

$x+y+z=0$

$\lambda x-3y=\mu$

System has an infinite number of Solutions then 

$\Delta =0,{\Delta }_1=0$

$$\begin{gathered} \Delta =0 {\Delta }_1=0 \\ \Rightarrow \left|\begin{array}{ccc}8& 1& 4\\ 1& 1& 1\\ \lambda & -3& 0\end{array}\right|=0 \Rightarrow \left|\begin{array}{ccc}-2& 1& 4\\ 0& 1& 1\\ \mu & -3& 0\end{array}\right|=0 \\ \Rightarrow \lambda \left(1-4\right)+3\left(8-4\right)=0 \Rightarrow \mu \left(1-4\right)+3\left(-2-0\right)=0 \\ \Rightarrow -3\lambda +12=0 \Rightarrow -3\mu -6=0 \\ \Rightarrow 3\lambda =12\Rightarrow \lambda =4 \Rightarrow -3\mu -6=0 \\ \Rightarrow -3\mu =6 \\ \Rightarrow \mu =-2 \end{gathered}$$

$po\mathrm{int}\left(\lambda ,\mu ,\frac{-1}{2}\right)=\left(4,-2,\frac{-1}{2}\right)$  

Distance $=\left|\frac{8\left(4\right)+\left(-2\right)+4\left(\frac{-1}{2}\right)+2}{\sqrt{64+1+16}}\right|=\frac{30}{9}=\frac{10}{3}$