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

$$
[\mathbf{a} \mathbf{b} \mathbf{c}]=1(6-3)-2(0-3)+0=3+6=9
$$