The lines $$x = ay - 1 = z - 2$$ and $$x = 3y - 2 = bz - 2$$, $$(ab \neq 0)$$ are coplanar, if:

We rewrite the two lines in symmetric form. For the first line $$x = ay - 1 = z - 2$$: set $$x = ay - 1$$ so $$y = \frac{x+1}{a}$$, and $$z - 2 = x$$. In parametric form with parameter $$t = x$$: the line passes through $$(-1, 0, 2)$$ (when $$t = 0$$, $$x = -1$$ does not work cleanly; let us use $$ay - 1 = x$$ and $$z - 2 = x$$).

Setting each expression equal to $$t$$: $$x = t$$, $$ay - 1 = t \Rightarrow y = \frac{t+1}{a}$$, $$z - 2 = t \Rightarrow z = t + 2$$. So direction vector of $$L_1$$ is $$\left(1, \frac{1}{a}, 1\right)$$, or equivalently $$(a, 1, a)$$, and a point on $$L_1$$ is $$A_1 = (0, \frac{1}{a}, 2)$$.

For the second line $$x = 3y - 2 = bz - 2$$: set each equal to $$s$$: $$x = s$$, $$3y - 2 = s \Rightarrow y = \frac{s+2}{3}$$, $$bz - 2 = s \Rightarrow z = \frac{s+2}{b}$$. Direction vector of $$L_2$$ is $$\left(1, \frac{1}{3}, \frac{1}{b}\right)$$, or $$(3b, b, 3)$$, and a point on $$L_2$$ is $$A_2 = (0, \frac{2}{3}, \frac{2}{b})$$.

Two lines are coplanar if and only if the scalar triple product $$[\overrightarrow{d_1}, \overrightarrow{d_2}, \overrightarrow{A_1A_2}] = 0$$, where $$\overrightarrow{d_1}$$ and $$\overrightarrow{d_2}$$ are direction vectors and $$\overrightarrow{A_1A_2}$$ connects a point on each line.

$$\overrightarrow{A_1A_2} = A_2 - A_1 = \left(0, \frac{2}{3} - \frac{1}{a}, \frac{2}{b} - 2\right)$$

The scalar triple product determinant is: $$\begin{vmatrix} 1 & \frac{1}{a} & 1 \\ 1 & \frac{1}{3} & \frac{1}{b} \\ 0 & \frac{2}{3}-\frac{1}{a} & \frac{2}{b}-2 \end{vmatrix} = 0$$

Expanding along the first row: $$1\left[\frac{1}{3}\left(\frac{2}{b}-2\right) - \frac{1}{b}\left(\frac{2}{3}-\frac{1}{a}\right)\right] - \frac{1}{a}\left[1\cdot\left(\frac{2}{b}-2\right) - \frac{1}{b}\cdot 0\right] + 1\left[1\cdot\left(\frac{2}{3}-\frac{1}{a}\right) - \frac{1}{3}\cdot 0\right] = 0$$

$$\left[\frac{2}{3b}-\frac{2}{3} - \frac{2}{3b}+\frac{1}{ab}\right] - \frac{1}{a}\left[\frac{2}{b}-2\right] + \left[\frac{2}{3}-\frac{1}{a}\right] = 0$$

$$\left[\frac{1}{ab}-\frac{2}{3}\right] - \frac{2}{ab}+\frac{2}{a} + \frac{2}{3}-\frac{1}{a} = 0$$

$$\frac{1}{ab} - \frac{2}{3} - \frac{2}{ab} + \frac{2}{a} + \frac{2}{3} - \frac{1}{a} = 0$$

$$-\frac{1}{ab} + \frac{1}{a} = 0 \implies \frac{1}{a}\left(1 - \frac{1}{b}\right) = 0$$

Since $$ab \neq 0$$, we have $$a \neq 0$$, so $$1 - \frac{1}{b} = 0$$, which gives $$b = 1$$. The value of $$a$$ can be any nonzero real number. Therefore, the condition is $$b = 1$$ and $$a \in \mathbb{R} - \{0\}$$.