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Let $\vec{a}, \vec{b}, \vec{c} $ be distinct non-negative numbers. If the vectors $a\hat{i}+a\hat{j}+c\hat{k}$ , $\hat{i}+\hat{k}$ and $c\hat{i}+c\hat{j}+b\hat{k}$ lie in a plane, then c is





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$\vec{a}=a\hat{i}+a\hat{j}+c\hat{k}\, ,\, \vec{b}=\hat{i}+\hat{k}\, \&\, \vec{c}=c\hat{i}+c\hat{j}+b\hat{k}$ are coplanar.

$\Rightarrow\begin{vmatrix}{a} & {a} & {c} \\ {1} & {0} & {1} \\ {c} & {c} & {b}\end{vmatrix}=0$

$\Rightarrow-ac-ab+ac+{c}^2=0$

$\Rightarrow{c}^2=ab$


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