$$\begin{aligned} \text{Let}\quad & \overrightarrow{\mathbf{c}}=x \hat{\mathbf{i}}+y \hat{\mathbf{j}}+z \hat{\mathbf{k}} \\ & \text{Also,}\quad \overrightarrow{\mathbf{a}}=\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}} \text { and } \overrightarrow{\mathbf{b}}=\hat{\mathbf{j}}-\hat{\mathbf{k}} \end{aligned}$$

For $\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{c}}=\overrightarrow{\mathbf{b}}$,

$$\begin{aligned} & \Rightarrow \quad\left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 1 & 1 & 1 \\ x & y & z \end{array}\right|=\hat{\mathbf{j}}-\hat{\mathbf{k}} \\ & \Rightarrow \quad \hat{\mathbf{i}}(z-y)-\hat{\mathbf{j}}(z-x)+\hat{\mathbf{k}}(y-x)=\hat{\mathbf{j}}-\hat{\mathbf{k}} \end{aligned}$$

$$ \begin{aligned} \therefore \quad & z-y=0 \quad\text{.... (i)}\\ & x-z=1 \quad\text{.... (ii)}\\ & x-y=1\quad\text{.... (iii)} \end{aligned}$$

$$\begin{aligned} \text{Also,}\quad \overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{c}} & =3 \\ (\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}) \cdot(x \hat{\mathbf{i}}+y \hat{\mathbf{j}}+z \mathbf{k}) & =3 \\ \Rightarrow\quad x+y+z & =3\quad \text{.... (iv)} \end{aligned}$$

On adding Eqs. (ii) and (iii), we get

$$2 x-y-z=2\quad\text{.... (v)}$$

$$\begin{aligned} &\text { On solving Eqs. (iv) and (v), we get }\\ &\begin{aligned} x & =\frac{5}{3} \\ \therefore\quad y & =\frac{5}{3}-1=\frac{2}{3} \text { and } z=\frac{2}{3} \\ \text{Now,}\quad \overrightarrow{\mathbf{c}} & =\frac{5}{3} \hat{\mathbf{i}}+\frac{2}{3} \hat{\mathbf{j}}+\frac{2}{3} \hat{\mathbf{k}} \\ & =\frac{1}{3}(5 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}) \end{aligned} \end{aligned}$$