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

The system of linear equations is:

$$8x + y + 4z = -2 \quad \cdots(1)$$

$$x + y + z = 0 \quad \cdots(2)$$

$$\lambda x - 3y = \mu \quad \cdots(3)$$

$$D = \begin{vmatrix} 8 & 1 & 4 \\ 1 & 1 & 1 \\ \lambda & -3 & 0 \end{vmatrix}$$

$$= 8(0 + 3) - 1(0 - \lambda) + 4(-3 - \lambda) = 24 + \lambda - 12 - 4\lambda = 12 - 3\lambda$$

Setting $$D = 0$$: $$\lambda = 4$$.

From $$(1) - 8 \times (2)$$: $$-7y - 4z = -2 \quad \cdots(4)$$

From (2): $$x = -y - z$$. Substituting into (3) with $$\lambda = 4$$:

$$4(-y - z) - 3y = \mu \implies -7y - 4z = \mu$$

Comparing with (4): $$\mu = -2$$.

Point: $$(\lambda, \mu, -\frac{1}{2}) = (4, -2, -\frac{1}{2})$$.

Plane: $$8x + y + 4z + 2 = 0$$.

$$d = \frac{|8(4) + 1(-2) + 4(-\frac{1}{2}) + 2|}{\sqrt{64 + 1 + 16}} = \frac{|32 - 2 - 2 + 2|}{\sqrt{81}} = \frac{30}{9} = \frac{10}{3}$$

Therefore, the correct answer is Option D: $$\dfrac{10}{3}$$.