If the system of linear equations
$$2x + 2ay + az = 0$$
$$2x + 3by + bz = 0$$
$$2x + 4cy + cz = 0$$,
where $$a, b, c \in R$$ are non-zero and distinct; has a non-zero solution, then

For a homogeneous system of linear equations to possess a nonzero solution, the determinant of its coefficient matrix must be exactly equal to zero. Let us set up the determinant of the coefficients for the variables $$x$$, $$y$$, and $$z$$. Expanding this determinant gives

$$\begin{vmatrix} 2 & 2a & a \\ 2 & 3b & b \\ 2 & 4c & c \end{vmatrix} = 0$$

We can simplify this calculation by factoring out the $$2$$ from the first column, yielding

$$2 \begin{vmatrix} 1 & 2a & a \\ 1 & 3b & b \\ 1 & 4c & c \end{vmatrix} = 0$$

To evaluate this efficiently, we apply row transformations. Subtracting the first row from both the second and third rows simplifies the matrix to

$$\begin{vmatrix} 1 & 2a & a \\ 0 & 3b - 2a & b - a \\ 0 & 4c - 2a & c - a \end{vmatrix} = 0$$

Expanding this determinant along the first column gives the algebraic equation

$$1 \cdot [(3b - 2a)(c - a) - (b - a)(4c - 2a)] = 0$$

Multiplying the terms inside the brackets results in

$$(3bc - 3ab - 2ac + 2a^2) - (4bc - 2ab - 4ac + 2a^2) = 0$$

Opening the parentheses and reversing the signs for the second group yields

$$3bc - 3ab - 2ac + 2a^2 - 4bc + 2ab + 4ac - 2a^2 = 0$$

Canceling the common terms and grouping the remaining variables provides the relation

$$-bc - ab + 2ac = 0$$

Moving the negative terms to the right side isolates the positive components to give

$$2ac = ab + bc$$

Since the problem states that $$a$$, $$b$$, and $$c$$ are nonzero real numbers, dividing the entire equation by the product $$abc$$ gives

$$\frac{2ac}{abc} = \frac{ab}{abc} + \frac{bc}{abc}$$

Simplifying the fractions reveals the final relationship

$$\frac{2}{b} = \frac{1}{c} + \frac{1}{a}$$

This resulting equation is the standard defining condition for three terms to be in an arithmetic progression. Therefore, the reciprocals of $$a$$, $$b$$, and $$c$$ form an arithmetic progression, making the first option the correct choice.

 the correct choice.