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 $\left( {\lambda ,\mu , - {1 \over 2}} \right)$ from the plane $8x + y + 4z + 2 = 0$ is :

<p>$$\Delta = \left| {\matrix{ 8 & 1 & 4 \cr 1 & 1 & 1 \cr \lambda & { - 3} & 0 \cr } } \right|$$</p> <p>$= 8(3) - 1( - \lambda ) + 4( - 3 - \lambda )$</p> <p>$= 24 + \lambda - 12 - 4\lambda$</p> <p>$= 12 - 3\lambda$</p> <p>So for $\lambda = 4$, it is having infinitely many solutions.</p> <p>$${\Delta _x} = \left| {\matrix{ { - 2} & 1 & 4 \cr 0 & 1 & 1 \cr \mu & { - 3} & 0 \cr } } \right|$$</p> <p>$= - 2(3) - 1( - \mu )+4( - \mu )$</p> <p>$= - 6 - 3\mu = 0$</p> <p>For $\mu = - 2$</p> <p>Distance of $(4, - 2,{{ - 1} \over 2})$ from $8x + y + 4z + 2 = 0$</p> <p>$= {{32 - 2 - 2 + 2} \over {\sqrt {64 + 1 + 16} }} = {{10} \over 3}$ units</p>