Given lines can be written as

$\frac{x+1}{a}=\frac{y-0}{1}=\frac{z-1}{a}$

$\frac{x+2}{3}=\frac{y-0}{1}=\frac{z-0}{{\displaystyle \frac{3}{b}}}$

We know that lines $\frac{x-x_1}{a_1}=\frac{y-y_1}{b_1}=\frac{z-z_1}{c_1}$ and $\frac{x-x_2}{a_2}=\frac{y-y_2}{b_2}=\frac{z-z_2}{c_2}$ are coplanar if $\left|\begin{array}{ccc}{x}_2-x_1& y_2-y_1& z_2-z_1\\ a_1& b_1& c_1\\ a_2& b_2& c_2\end{array}\right|=0$.

Hence,

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

where, $b\ne 0\Rightarrow a\ne 0$, since $ab\ne 0$.

$\Rightarrow -\left(\frac{3}{ b}-a\right)-1\left(a-3\right)=0$

$\Rightarrow a-\frac{3}{b}-a+3=0$

$\Rightarrow b=1$

$\therefore b=1,a\in R-\left\{0\right\}$