Given vectors \(\mathbf{a}=3 \hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}}, \mathbf{b}=\hat{\mathbf{i}}-4 \hat{\mathbf{j}}+5 \hat{\mathbf{k}}\) and \(\mathbf{c}=4 \hat{\mathbf{i}}+5 \hat{\mathbf{j}}-\hat{\mathbf{k}}\)
\(\begin{aligned}
& \because \mathbf{b} \times \mathbf{c}= \left|\begin{array}{ccc}
\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\
1 & -4 & 5 \\
4 & 5 & -1
\end{array} \right|=\hat{\mathbf{i}}(4-25)-\hat{\mathbf{j}}(-\mathbf{l}-20) +\hat{\mathbf{k}}(5+16) \\
&=-21 \hat{\mathbf{i}}+21 \hat{\mathbf{j}}+21 \hat{\mathbf{k}}=21(-\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}) \\
& \Rightarrow|\mathbf{b} \times \mathbf{c}|=21 \sqrt{3}
\end{aligned}\)
Now, vector \(\mathbf{r}\), which is perpendicular to \(b\) and \(\mathbf{c}\), so
\(\begin{aligned}
\mathbf{r}= &\pm|\mathbf{r}| \frac{\mathbf{b} \times \mathbf{c}}{|\mathbf{b} \times \mathbf{c}|} \\
\because & \mathbf{r} \cdot \mathbf{a}=9 \\
\Rightarrow & \pm \frac{|\mathbf{r}|}{|\mathbf{b} \times \mathbf{c}|}[21(-\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}) \cdot(3 \hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}})]=9 \\
\Rightarrow & \pm \frac{|\mathbf{r}|}{21 \sqrt{3}}[21(-3+1-1)]=9 \\
\Rightarrow & |\mathbf{r}|=3 \sqrt{3} \quad (\because|\mathbf{r}| > 0)\\
\therefore & \mathbf{r}= \pm 3 \sqrt{3} \frac{21(-\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}})}{21 \sqrt{3}}= \pm 3(-\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}) \\
& =-3(\hat{\mathbf{i}}+\hat{\mathbf{j}}+\mathbf{k}) \text { or } 3(\hat{\mathbf{i}}-\hat{\mathbf{j}}-\hat{\mathbf{k}})
\end{aligned}\)
Hence,option (a) is correct.