If $\overline{\mathrm{a}}\times \overline{\mathrm{r}}=\overline{\mathrm{b}}+\mathrm{\lambda }\overline{\mathrm{a}}, \overline{\mathrm{a}}·\overline{\mathrm{r}}=3$ where $\overline{\mathrm{a}}=2\overline{\mathrm{i}}+\overline{\mathrm{j}}-\overline{\mathrm{k}}, \overline{\mathrm{b}}=-\overline{\mathrm{i}}-2\overline{\mathrm{j}}+\overline{\mathrm{k}}$ then $\mathrm{\lambda }=$

$\begin{array}{rcl}\overline{\mathrm{a}}\times (\overline{\mathrm{a}}\times \overline{\mathrm{r}})& =& \overline{\mathrm{a}}\times \overline{\mathrm{b}}+0\\ \left(\overline{\mathrm{a}}.\overline{r}\right)\overline{\mathrm{a}}-\left(\overline{\mathrm{a}}.\overline{\mathrm{a}}\right)\overline{r}& =& \overline{\mathrm{a}}\times \overline{\mathrm{b}}\\ 3\overline{\mathrm{a}}-6\overline{r}& =& \overline{\mathrm{a}}\times \overline{\mathrm{b}}\\ 6\overline{r}& =& 3\overline{\mathrm{a}}-\left(\overline{\mathrm{a}}\times \overline{\mathrm{b}}\right)\end{array}$

$\begin{array}{rcl}\overline{\mathrm{a}}\times \overline{\mathrm{b}} & =& \left|\begin{array}{ccc}\mathrm{i}& \mathrm{j}& \mathrm{k}\\ 2& 1& -1\\ -1& -2& 1\end{array}\right|\end{array}$

$\begin{array}{rcl} & =& \mathrm{i}(1-2)-\mathrm{j}(2-1)+\mathrm{k}(-4+1)\end{array}$

$\begin{array}{rcl} & =& -\mathrm{i}-\mathrm{j}-3\mathrm{k}\end{array}$

$\begin{array}{rcl}6\overline{r}& =& 6\mathrm{i}+3\mathrm{j}-3\mathrm{k}+\mathrm{i}+\mathrm{j}+3\mathrm{k}\end{array}$

$\begin{array}{rcl} & & \overline{r}\end{array}=\frac{7\overline{i}}{6}+\frac{4\overline{j}}{6}$

$\begin{array}{rcl}\overline{\mathrm{a}}\times \overline{\mathrm{r}}& =& \left|\begin{array}{ccc}\mathrm{i}& \mathrm{j}& \mathrm{k}\\ 2& 1& -1\\ \frac{7}{6}& \frac{4}{6}& 0\end{array}\right|=\mathrm{i}\left(\frac{4}{6}\right)-\mathrm{j}\left(\frac{7}{6}\right)+\mathrm{k}\left(\frac{8-7}{6}\right)\\ & =& \frac{4i-7j+k}{6}\end{array}$

$\begin{array}{rcl}\frac{4i-7j+k}{6}& =& -\mathrm{i}-2\mathrm{j}+\mathrm{k}+\mathrm{\lambda } (2\mathrm{i}+\mathrm{j}-\mathrm{k}) \end{array}$

$\begin{array}{rcl}\frac{4}{6}& =& -1+2\mathrm{\lambda } \Rightarrow \mathrm{\lambda }=\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$6$}\right. \end{array}$