\(\overline{\mathrm{a}}=2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}, \quad \overline{\mathrm{b}}=\hat{\mathrm{i}}-2 \hat{\mathrm{j}}+\hat{\mathrm{k}}, \quad \overline{\mathrm{c}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}}\) are three vectors. For a vector \(\overline{\mathrm{r}}\) with \(\overline{\mathrm{r}} \times \overline{\mathrm{a}}=\overline{\mathrm{b}}\) and \(\overline{\mathrm{r}} \cdot \overline{\mathrm{c}}=3,|\overline{\mathrm{r}}|\) is

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