Let $\mathbf{b}=x \hat{i}+y \hat{j}+z \hat{k}$
If $\mathbf{a} \times \mathbf{b}=\mathbf{c}$, then $\mathbf{c}$ is perpendicular to both $\mathbf{a}$ and $\mathbf{b}$
$$
\begin{aligned}
\mathbf{b} \cdot \mathbf{c} & =0 \\
y-z & =0 \\
y & =z...(i)
\end{aligned}
$$
Also $\mathbf{a} \cdot \mathbf{b}=3$
$$
\begin{array}{r}
x+y+z=3 \\
x+2 y=3...(ii)
\end{array}
$$
$$
\mathbf{a} \times \mathbf{b}=\mathbf{c}
$$
$$
\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
1 & 1 & 1 \\
x & y & z
\end{array}\right|=\hat{j}-\hat{k}
$$
$$
\begin{aligned}
& \hat{i}(z-y)-\hat{j}(z-x)+\hat{k}(y-x)=\hat{j}-\hat{k} \\
& \Rightarrow x-z=1 \text { and } x-y=1...(ii)
\end{aligned}
$$
Subtracting Eq. (ii) from Eq. (i),
$$
3 y=2 \Rightarrow y=\frac{2}{3}
$$
Also, $z=\frac{2}{3}$ [from Eq. (i)]
$$
\begin{array}{cc}
& x+y+z=3 \\
\Rightarrow & x+\frac{4}{3}=3 \Rightarrow x=\frac{5}{3} \\
\therefore & \mathbf{b}=\frac{1}{3}(5 \hat{i}+2 \hat{j}+2 \hat{k})
\end{array}
$$