Lines are $$x = ay - 1 = z - 2$$

$$\therefore$$ $${x \over 1} = {{y - {1 \over a}} \over {{1 \over a}}} = {{z - 2} \over 1}$$ .... (i)

and $$x = 3y - 2 = bz - 2$$

$$\therefore$$ $${x \over 1} = {{y - {2 \over 3}} \over {{1 \over 3}}} = {{z - {2 \over b}} \over {{1 \over b}}}$$ .... (ii)

$$\therefore$$ lines are co-planar

$$\therefore$$ $$\left| {\matrix{ 0 & { - {1 \over a} + {2 \over 3}} & { - 2 + {2 \over b}} \cr 1 & {{1 \over a}} & 1 \cr 1 & {{1 \over 3}} & {{1 \over b}} \cr } } \right| = 0$$

$$\therefore$$ $$\left| {\matrix{ 0 & {{2 \over 3} - {1 \over a}} & {{2 \over b} - 2} \cr 0 & {{1 \over a} - {1 \over 3}} & {1 - {1 \over b}} \cr 1 & {{1 \over 3}} & {{1 \over b}} \cr } } \right| = 0$$

$$\therefore$$ $${1 \over a} - {1 \over {ab}} = 0$$

$$\Rightarrow$$ b = 1 and a $$\in$$ R $$-$$ {0}