$$\Delta = \left| {\matrix{ 8 & 1 & 4 \cr 1 & 1 & 1 \cr \lambda & { - 3} & 0 \cr } } \right|$$

$$ = 8(3) - 1( - \lambda ) + 4( - 3 - \lambda )$$

$$ = 24 + \lambda - 12 - 4\lambda $$

$$ = 12 - 3\lambda $$

So for $$\lambda = 4$$, it is having infinitely many solutions.

$${\Delta _x} = \left| {\matrix{ { - 2} & 1 & 4 \cr 0 & 1 & 1 \cr \mu & { - 3} & 0 \cr } } \right|$$

$$ = - 2(3) - 1( - \mu )+4( - \mu )$$

$$ = - 6 - 3\mu = 0$$

For $$\mu = - 2$$

Distance of $$(4, - 2,{{ - 1} \over 2})$$ from $$8x + y + 4z + 2 = 0$$

$$ = {{32 - 2 - 2 + 2} \over {\sqrt {64 + 1 + 16} }} = {{10} \over 3}$$ units