If the system of linear equations. 8x + y + 4z = - 2 x + y + z = 0...

Given the system of linear equations.

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

$x + y + z = 0$

$\lambdax - 3y = \mu$

has infinitely many solutions, so

D = \left|\begin{cases}8 & 1 & 4 \\ 1 & 1 & 1 \\ \lambda & - 3 & 0\end{cases}\right| = 0 ⇒ \lambda = 4

Also ${D}_{1} = {D}_{2} = {D}_{3} = 0$

{D}_{1} = \left|\begin{cases}- 2 & 1 & 4 \\ 0 & 1 & 1 \\ \mu & - 3 & 0\end{cases}\right| = 0 ⇒ - 6 - 3\mu = 0

So $\mu = - 2$

Hence the point will be \left(4, - 2, - \dfrac{1}{2}\right)$</p><p>Now distance of point \left(4, - 2, - \dfrac{1}{2}\right)$ from plane $8x + y + 4z + 2 = 0$

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