Given

$\overrightarrow{a}=\hat{i}+\hat{j}+\hat{k}$

$\overrightarrow{b}=2\hat{i}-\hat{j}+3\hat{k}$

$\overrightarrow{c}=\hat{i}-\hat{j}$

Now, $\overrightarrow{a}\times \overrightarrow{b}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 1& 1\\ 2& -1& 3\end{array}\right|$

          $=4\hat{i}-\hat{j}-3\hat{k}$

$\overrightarrow{b}\times \overrightarrow{c}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& -1& 3\\ 1& -1& 0\end{array}\right|$

           $=3\hat{i}+3\hat{j}-\hat{k}$

Similarly, $\overrightarrow{c}\times \overrightarrow{a}=-\hat{i}-\hat{j}+2\hat{k}$

Now we have

$6\hat{i}+2\hat{j}+3\hat{k}={\lambda }_1(\overrightarrow{a}\times \overrightarrow{b})+{\lambda }_2(\overrightarrow{b}\times \overrightarrow{c})+{\lambda }_3(\overrightarrow{c}\times \overrightarrow{a})$

$={\lambda }_1\left(4\hat{i}-\hat{j}-3\hat{k}\right)+{\lambda }_2\left(3\hat{i}+3\hat{j}-\hat{k}\right)+{\lambda }_3\left(-\hat{i}-\hat{j}+2\hat{k}\right)$

$=\left(4{\lambda }_1+3{\lambda }_2-{\lambda }_3\right)\hat{i}+\left(-{\lambda }_1+3{\lambda }_2-{\lambda }_3\right)\hat{j}+\left(-3{\lambda }_1-{\lambda }_2+2{\lambda }_3\right)\hat{k}$

On comparing, we get

$4{\lambda }_1+3{\lambda }_2-{\lambda }_3=6 ...\left(\mathrm{i}\right)$

$-{\lambda }_1+3{\lambda }_2-{\lambda }_3=2 ...\left(\mathrm{ii}\right)$

$-3{\lambda }_1-{\lambda }_2+2{\lambda }_3=3 ...\left(\mathrm{iii}\right)$

On solving, we get

$\left({\lambda }_1,{\lambda }_2,{\lambda }_3\right)\equiv \left(\frac{4}{5},\frac{11}{5},\frac{19}{5}\right)$