$\frac{\mathrm{x}-0}{1}=\frac{\mathrm{y}-\frac{1}{\mathrm{a}}}{\frac{1}{\mathrm{a}}}=\frac{\mathrm{z}-2}{1}$                                           ….(1)

$\frac{\mathrm{x}-0}{1}=\frac{\mathrm{y}-\frac{2}{3}}{\frac{1}{3}}=\frac{\mathrm{z}-\frac{2}{\mathrm{b}}}{\frac{1}{\mathrm{b}}}$                                         ….(2)

For coplanar :

$\left|\begin{array}{ccc}{\mathrm{a}}_2-{\mathrm{a}}_1& {\mathrm{b}}_2-{\mathrm{b}}_1& {\mathrm{c}}_2-{\mathrm{c}}_1\\ {\mathrm{\ell }}_1& {\mathrm{m}}_1& {\mathrm{n}}_1\\ {\mathrm{\ell }}_2& {\mathrm{m}}_2& {\mathrm{n}}_2\end{array}\right|=0$

$\left|\begin{array}{ccc}0& \frac{2}{3}-\frac{1}{\mathrm{a}}& \frac{2}{\mathrm{b}}-2\\ 1& \frac{1}{\mathrm{a}}& 1\\ 1& \frac{1}{3}& \frac{1}{\mathrm{b}}\end{array}\right|=0$

$\Rightarrow -(\frac{2}{3}-\frac{1}{\mathrm{a}})\{\frac{1}{\mathrm{b}}-1\}+(\frac{2}{\mathrm{b}}-2)\{\frac{1}{3}-\frac{1}{\mathrm{a}}\}=0$

$\Rightarrow (\frac{1}{\mathrm{a}}-\frac{2}{3})(\frac{1}{\mathrm{b}}-1)+(\frac{2}{\mathrm{b}}-2)(\frac{1}{3}-\frac{1}{\mathrm{a}})=0$

⇒  (3 – 2a)(1 – b) + (2 – 2b)(a – 3) = 0

⇒ 3 – 3b – 2a + 2ab + 2a – 6 – 2ab + 6b = 0

⇒ 3b – 3 = 0

⇒ b = 1, a $\in$ R – {0}