Let $\overrightarrow{\mathrm{a}}=2 \hat{\mathrm{i}}+\lambda_{1} \hat{\mathrm{j}}+3 \hat{\mathrm{k}}, \overrightarrow{\mathrm{b}}=4 \hat{\mathrm{i}}+\left(3-\lambda_{2}\right) \hat{\mathrm{j}}+6 \hat{\mathrm{k}}$ and $\overrightarrow{\mathrm{c}}=3 \hat{\mathrm{i}}+6 \hat{\mathrm{j}}+\left(\lambda_{3}-1\right) \hat{\mathrm{k}}$ be three vectors such that $\overrightarrow{\mathrm{b}}=2 \overrightarrow{\mathrm{a}}$ and $\overrightarrow{\mathrm{a}}$ is perpendicular to $\overrightarrow{\mathrm{c}}$. Then a possible value of $\left(\lambda_{1}, \lambda_{2}, \lambda_{3}\right)$ is :-
Correct Option: , 2
$4 \hat{\mathrm{i}}+\left(3-\lambda_{2}\right) \hat{\mathrm{j}}+6 \hat{\mathrm{k}}=4 \hat{\mathrm{i}}+2 \lambda_{1} \hat{\mathrm{j}}+6 \hat{\mathrm{k}}$
$\Rightarrow 3-\lambda_{2}=2 \lambda_{1} \Rightarrow 2 \lambda_{1}+\lambda_{2}=3$ ..........(1)
Given $\vec{a} \cdot \vec{c}=0$
$\Rightarrow 6+6 \lambda_{1}+3\left(\lambda_{3}-1\right)=0$
$\Rightarrow 2 \lambda_{1}+\lambda_{3}=-1$ ....................(2)
Now $\left(\lambda_{1}, \lambda_{2}, \lambda_{3}\right)=\left(\lambda_{1}, 3-2 \lambda_{1},-1-2 \lambda_{1}\right)$
Now check the options, option (2) is correct