Given, $\quad \mathbf{a}=x \hat{\mathbf{i}}+2 \hat{\mathbf{j}}, \quad \mathbf{b}=y \hat{\mathbf{j}}+3 \mathbf{k} \quad$ and $\mathbf{c}=x \hat{\mathbf{i}}+y \hat{\mathbf{j}}+z \hat{\mathbf{k}}$
$$
\begin{aligned}
& \text { Now, } \quad \mathbf{a} \times \mathbf{b}=\left|\begin{array}{ccc}
\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\
x & 2 & 0 \\
0 & y & 3
\end{array}\right| \\
&=\hat{\mathbf{i}}(6-0)-\hat{\mathbf{j}}(3 x-0)+\hat{\mathbf{k}}(x y-0) \\
&=6 \hat{\mathbf{i}}-3 x \hat{\mathbf{j}}+x y \hat{\mathbf{k}} \\
& \Rightarrow 6 \hat{\mathbf{i}}-3 x \hat{\mathbf{j}}+x y \hat{\mathbf{k}}=3 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+\hat{\mathbf{k}}
\end{aligned}
$$
On equating the coefficients of $\mathbf{i}$, $\mathbf{j}$ and $\mathbf{k}$, we get
$$
\begin{aligned}
& \quad z=3, x=1 \text { and } x y=1 \\
& \therefore \quad x y=1 \Rightarrow y=1 \\
& \therefore \quad a=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}, \mathbf{b}=\hat{\mathbf{j}}+3 \hat{\mathbf{k}} \\
& \text { and } \quad \quad \mathbf{c}=\hat{\mathbf{i}}+\hat{\mathbf{j}}+6
\end{aligned}
$$