(d) Three vectors $\mathrm{a}, \mathrm{b}$ and $\mathrm{c}$ are given as, $a=2 \hat{i}+\hat{k}, b=\hat{i}+\hat{j}+\hat{k}$ and $c=4 \hat{i}-3 \hat{j}+7 \hat{k}$
Given condition, $\mathrm{r} \times \mathrm{b}=\mathrm{c} \times \mathrm{b}$
$\begin{aligned}
& \mathrm{r} \times \mathrm{b}-\mathrm{c} \times \mathrm{b}=0 \\
& (\mathrm{r}-\mathrm{c}) \times \mathrm{b}=0
\end{aligned}$
It means, $(\mathrm{r}-\mathrm{c}) \| \mathrm{b}$
So, $r-c=\lambda b$


Also given, $\mathrm{r} \cdot \mathrm{a}=0$
$(c+\lambda b) \cdot a=0$ [using Eq. (i)]
$\begin{aligned}
& (4 \hat{i}-3 \hat{j}+7 \hat{k}+\lambda \hat{i}+\lambda \hat{j}+\lambda \hat{k}) \cdot(2 \hat{i}+\hat{k})=0 \\
& {[(4+\lambda) \cdot \hat{i}+(-3+\lambda) \hat{j}+(7+\lambda) \hat{k}] \cdot(2 \hat{i}+\hat{k})=0} \\
& (4+\lambda) \cdot 2+(7+\lambda) \cdot 1=0 \\
& 8+2 \lambda+7+\lambda=0 \\
& 3 \lambda=-15 \\
& \therefore \quad \lambda=-5
\end{aligned}$
Put the value of $\lambda$ in Eq. (i), we get
$\begin{aligned}
& r=4 \hat{i}-3 \hat{j}+7 \hat{k}-5(\hat{i}+\hat{j}+\hat{k}) \\
& =4 \hat{i}-3 \hat{j}+7 \hat{k}-5 \hat{i}-5 \hat{j}-5 \hat{k}=-\hat{i}-8 \hat{j}+2 \hat{k}
\end{aligned}$