$\begin{aligned} & \text { Let } \overline{\mathrm{r}}=x \hat{\mathrm{i}}+y \hat{\mathrm{j}}+\mathrm{z} \hat{\mathrm{k}} \\ & \overline{\mathrm{r}} \times \overline{\mathrm{a}}=\overline{\mathrm{b}} \\ & \Rightarrow\left|\begin{array}{llr}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ x & y & \mathrm{z} \\ 2 & 3 & 4\end{array}\right|=\overline{\mathrm{b}} \\ & \Rightarrow(4 y-3 \mathrm{z}) \hat{\mathrm{i}}-(4 x-2 \mathrm{z}) \hat{\mathrm{j}}+(3 x-2 y) \hat{\mathrm{k}}=\hat{\mathrm{i}}-2 \hat{\mathrm{j}}+\hat{\mathrm{k}}\end{aligned}$
$\left.\begin{array}{r}4 y-3 z=1 \\ 4 x-2 z=2 \\ 3 x-2 y=1\end{array}\right)$ ...(i)
$\begin{aligned} & \overline{\mathrm{r}} \cdot \overline{\mathrm{c}}=3 \\ & \Rightarrow(x \hat{\mathrm{i}}+y \hat{\mathrm{j}}+\mathrm{z} \hat{\mathrm{k}}) \cdot(\hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}})=3\end{aligned}$
$\Rightarrow x+y-z=3$ ...(ii)
$x=5, y=7, z=9$
$\begin{aligned} \therefore \quad \overline{\mathrm{r}}= & 5 \hat{\mathrm{i}}+7 \hat{\mathrm{j}}+9 \hat{\mathrm{k}} \\ \Rightarrow|\overline{\mathrm{r}}| & =\sqrt{5^2+7^2+9^2} \\ & =\sqrt{155}\end{aligned}$