Let $\mathbf{a}=\hat{\mathbf{i}}+\hat{\mathbf{j}}+\mathbf{k}, \mathbf{b}=2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+3 \hat{\mathbf{k}}$ and $\mathbf{c}=\hat{\mathbf{i}}-\hat{\mathbf{j}}$
$\begin{aligned} \text { Now, } \mathbf{a} \times \mathbf{b} & =\left|\begin{array}{ccc}\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 1 & 1 & 1 \\ 2 & -1 & 3\end{array}\right| \\ & =\hat{\mathbf{i}}(3+1)-\hat{\mathbf{j}}(3-2)+\hat{\mathbf{k}}(-1-2)=4 \hat{\mathbf{i}}-\hat{\mathbf{j}}-3 \hat{\mathbf{k}}\end{aligned}$
and $\mathbf{b} \times \mathbf{c}=\left|\begin{array}{ccc}\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 2 & -1 & 3 \\ 1 & -1 & 0\end{array}\right|$ $=\hat{\mathbf{i}}(0+3)-\hat{\mathbf{j}}(0-3)+\hat{\mathbf{k}}(-2+1)=3 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-\hat{\mathbf{k}}$
and $\mathbf{c} \times \mathbf{a}=\left|\begin{array}{ccc}\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 1 & -1 & 0 \\ 1 & 1 & 1\end{array}\right|$
$$
=\hat{\mathbf{i}}(-1-0)-\hat{\mathbf{j}}(1-0)+\hat{\mathbf{k}}(1+1)=\hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{k}}
$$
Now, $6 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}=\lambda_1(\mathbf{a} \times \mathbf{b})+\lambda_2(\mathbf{b} \times \mathbf{c})$
$$
\begin{aligned}
\Rightarrow 6 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}= & \lambda_1(4 \hat{\mathbf{i}}-\hat{\mathbf{j}}-3 \hat{\mathbf{k}}) \quad+\lambda_3(\mathbf{c} \times \mathbf{a}) \\
& \lambda_2(3 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-\hat{\mathbf{k}})+\lambda_3(-\hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{k}}) \\
\Rightarrow 6 \hat{\mathbf{i}}+2 & \hat{\mathbf{j}}+3 \hat{\mathbf{k}}=\hat{\mathbf{i}}\left(4 \lambda_1+3 \lambda_2-\lambda_3\right) \\
& +\hat{\mathbf{j}}\left(-\lambda_1+3 \lambda_2-\lambda_3\right)+\hat{\mathbf{k}}\left(-3 \lambda_1-\lambda_2+2 \lambda_3\right)
\end{aligned}
$$
by comparing


Eqs. (i) and (ii), we get
$$
5 \lambda_1=4 \Rightarrow \lambda_1=\frac{4}{5}
$$
put in Eqs. (ii) and (iii), we get
$$
3 \lambda_2-\lambda_3=2+\frac{4}{5}
$$

$$
\text { and }-\lambda_2+2 \lambda_3=3+3 \cdot \frac{4}{5} \Rightarrow-\lambda_2+2 \lambda_3=\frac{27}{5}
$$
multiply by 3 .

Add Eqs. (iv) and (v)
$$
\begin{aligned}
& 5 \lambda_3=\frac{95}{5} \Rightarrow 5 \lambda_3=19 \\
& \lambda_3=\frac{19}{5}
\end{aligned}
$$
put in Eq. (iv)
$$
\begin{aligned}
3 \lambda_2 & =\frac{14}{5}+\frac{19}{5} \Rightarrow \lambda_2=\frac{33}{15} \\
\lambda_2 & =\frac{11}{5}=\left(\lambda_1, \lambda_2, \lambda_3\right)=\left(\frac{4}{5}, \frac{11}{5}, \frac{19}{5}\right)
\end{aligned}
$$