Let \(\mathbf{a}=-\hat{\mathbf{i}}-\hat{\mathbf{k}}, \mathbf{b}=-\hat{\mathbf{i}}+\hat{\mathbf{j}}\) and \(\mathbf{c}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}\) be three given vectors. If \(\mathbf{r}\) is a vector such that \(\mathbf{r} \times \mathbf{b}=\mathbf{c} \times \mathbf{b}\) and \(\mathbf{r} \cdot \mathbf{a}=0\), then the value of \(\mathbf{r} \cdot \mathbf{b}\) is

\[
\begin{aligned}
& \mathbf{r} \times \mathbf{b}=\mathbf{c} \times \mathbf{b} \\
& \Rightarrow \quad(\mathbf{r}-\mathbf{c}) \times \mathbf{b}=0 \Rightarrow \mathbf{r}-\mathbf{c}+\lambda \mathbf{b} \\
& \text { or } \quad \mathbf{r}=\mathbf{c}+\lambda \mathbf{b} \\
&
\end{aligned}
\]
Given, \(\mathbf{r} \cdot \mathbf{a}=0\), taking dot product with a for Eq. (i).
Now, \(\mathbf{r} \cdot \mathbf{a}=\mathbf{a} \cdot \mathbf{c}+\lambda \mathbf{a} \cdot \mathbf{b}\)
\(\therefore \quad \lambda=\frac{-\vec{a} \cdot \vec{c}}{\vec{a} \cdot \vec{b}} \quad[\because \vec{r} \cdot \vec{a}=0] \ldots\) (i)
From Eqs. (i) and (ii), we get
\[
\mathbf{r}=\mathbf{c}-\frac{\mathbf{a} \cdot \mathbf{c}}{\mathbf{a} \cdot \mathbf{b}} \mathbf{b}
\]
Taking dot with \(\mathbf{b}\), we get
\[
\mathbf{r} \cdot \mathbf{b}=\mathbf{c} \cdot \mathbf{b}-\frac{\mathbf{a} \cdot \mathbf{c}}{\mathbf{a} \cdot \mathbf{b}}(\mathbf{b} \cdot \mathbf{b})
\]

where, \(\left[\begin{array}{l}\mathbf{a}=-\hat{\mathbf{i}}-\hat{\mathbf{k}} \\ \mathbf{b}=-\hat{\mathbf{i}}+\hat{\mathbf{j}} \\ \mathbf{c}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}\end{array}\right]\)
\[
=(-1+2)-\frac{(-1-3)}{(1)}(1+1)=1+8=9
\]