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

We start with the homogeneous system

$$\begin{aligned} 2x+2ay+az&=0,\\ 2x+3by+bz&=0,\\ 2x+4cy+cz&=0, \end{aligned}$$

where $$a,\;b,\;c$$ are real, non-zero and pairwise distinct. A homogeneous system has a non-zero (non-trivial) solution iff the determinant of its coefficient matrix vanishes. Writing the coefficient matrix and its determinant, we have

$$\Delta=\begin{vmatrix} 2&2a&a\\[2pt] 2&3b&b\\[2pt] 2&4c&c \end{vmatrix}=0.$$

Every entry of the first column equals $$2$$, so we factor this common number out of the determinant:

$$\Delta =2\begin{vmatrix} 1&2a&a\\[2pt] 1&3b&b\\[2pt] 1&4c&c \end{vmatrix}=0.$$ Because $$2\neq0$$, the inner $$3\times3$$ determinant itself must be zero. Denote it by

$$D=\begin{vmatrix} 1&2a&a\\[2pt] 1&3b&b\\[2pt] 1&4c&c \end{vmatrix}=0.$$

To evaluate $$D$$ we perform the row operations
$$R_2\leftarrow R_2-R_1,\qquad R_3\leftarrow R_3-R_1.$$
This preserves the value of the determinant and produces

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

The first column now has exactly one non-zero entry (the top one). Expanding along this column gives

$$D=1\cdot\begin{vmatrix} 3b-2a&b-a\\[2pt] 4c-2a&c-a \end{vmatrix}=0.$$

So we require

$$\bigl(3b-2a\bigr)(c-a)-(b-a)\bigl(4c-2a\bigr)=0.$$

We expand each product term by term:

$$\begin{aligned} (3b-2a)(c-a)&=3bc-3ab-2ac+2a^2,\\ (b-a)(4c-2a)&=4bc-2ab-4ac+2a^2. \end{aligned}$$

Substituting these expressions we obtain

$$\bigl[3bc-3ab-2ac+2a^2\bigr]-\bigl[4bc-2ab-4ac+2a^2\bigr]=0.$$

Removing the brackets and combining like terms gives

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

which simplifies to

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

Multiplying the entire equation by $$-1$$ for convenience, we get

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

Because $$a,\;b,\;c\neq0$$, we divide every term by $$abc$$ to obtain

$$\frac{bc}{abc}+\frac{ab}{abc}-\frac{2ac}{abc}=0\;\Longrightarrow\; \frac1c+\frac1a-\frac{2}{b}=0.$$

Rearranging the terms yields

$$\frac{2}{b}=\frac1a+\frac1c.$$ Dividing both sides by $$2$$ gives

$$\frac1b=\frac12\left(\frac1a+\frac1c\right).$$

The middle quantity $$\dfrac1b$$ is now the arithmetic mean of $$\dfrac1a$$ and $$\dfrac1c$$. Thus $$\dfrac1a,\;\dfrac1b,\;\dfrac1c$$ form an arithmetic progression.

Therefore, the condition for the system to possess a non-trivial solution is precisely that the reciprocals $$\dfrac1a,\dfrac1b,\dfrac1c$$ are in A.P.

Hence, the correct answer is Option A.