Let a → = 2 i ^ + λ 1 j ^ + 3 k ^ ,   b → = 4 i ^ + 3 - λ 2 j ^ + 6 k ^ and c…

Since $\overrightarrow{b}=2\overrightarrow{a},$ so $3-{\lambda }_2=2{\lambda }_1$

${\lambda }_2=3-2{\lambda }_1 ...\left(1\right)$

We know that if two vectors $a_1\hat{i}+b_1\hat{j}+c_1\hat{k}$ and $a_2\hat{i}+b_2\hat{j}+c_2\hat{k}$ are perpendicular, then $a_1{a}_2+b_1{b}_2+c_1{c}_2=0$

Since, $\overrightarrow{a}$ is perpendicular to $\overrightarrow{c}$ so

$6+6{\lambda }_1+3\left({\lambda }_3-1\right)=0$

$\Rightarrow 6+6{\lambda }_1+3{\lambda }_3-3=0$

 $\Rightarrow {\lambda }_3=-1-2{\lambda }_1 ...\left(2\right)$

From equations $\left(1\right)$ and $\left(2\right),$ we get

$\left({\lambda }_1, {\lambda }_2, {\lambda }_3\right)=\left({\lambda }_1, 3-2{\lambda }_1, -1-2{\lambda }_1\right)$ where ${\lambda }_1\in R$

$\Rightarrow \left(-\frac{1}{2}, 4, 0\right)$ satisfies the above triplet.