Let $$\overrightarrow{\mathrm{a}}=\hat{i}+2 \hat{j}+3 \hat{k}, \overrightarrow{\mathrm{b}}=2 \hat{i}+3 \hat{j}-5 \hat{k}$$ and $\overrightarrow{\mathrm{c}}=3 \hat{i}-\hat{j}+\lambda \hat{k}$ be three vectors. Let $\overrightarrow{\mathrm{r}}$ be a unit vector along $\vec{b}+\vec{c}$. If $\vec{r} \cdot \vec{a}=3$, then $3 \lambda$ is equal to:

<p>$$\begin{aligned} & \vec{a}=\hat{i}+2 \hat{j}+3 \hat{k} \\ & \vec{b}=2 \hat{i}+3 \hat{j}-5 \hat{k} \\ & \vec{c}=3 \hat{i}-\hat{j}+\lambda \hat{k} \\ & \vec{b}+\vec{c}=5 \hat{i}+2 \hat{j}+(\lambda-5) \hat{k} \end{aligned}$$</p> <p>$\vec{r}$ is a unit vector along $\vec{b}+\vec{c}$</p> <p>$$\therefore \quad \vec{r}=\frac{5 \hat{i}+2 \hat{j}+(\lambda-5) \hat{k}}{\sqrt{25+4+(\lambda-5)^2}}$$</p> <p>Now, $\vec{r} \cdot \vec{a}=3$</p> <p>$\frac{1}{\sqrt{29+(\lambda-5)^2}}[5+4+3(\lambda-5)]=3$</p> <p>Squaring both sides</p> <p>$$\begin{aligned} & \Rightarrow \quad \frac{1}{29+(\lambda-5)^2}\left[9+3(\lambda-5)^2\right]=9 \\ & \Rightarrow \quad[3+(\lambda-5)]^2=29+(\lambda-5)^2 \\ & \Rightarrow \quad 9+(\lambda-5)^2+6(\lambda-5)=29+(\lambda-5)^2 \\ & \Rightarrow \quad 9+6(\lambda-5)=29 \\ & \Rightarrow \quad \lambda=\frac{20}{6}+5=\frac{25}{3} \\ & \therefore \quad 3 \lambda=3 \times \frac{25}{3}=25 \end{aligned}$$</p>