#### Let . Let a vector be in the plane containing . If is perpendicular to the vector and its projection on is 19 units, then is equal to___________

$Let\, \vec{\vartheta }= x\vec{a}+y\vec{b}$
$\vec{\vartheta }\cdot \left ( 3\hat{i}+2\hat{j}-\hat{k} \right )= 0$
$And\: \vec{\vartheta }\cdot \hat{a}= 19$
$Let\,\: 3\hat{i}+2\hat{j}-\hat{k}= \vec{c}$

$Let\,\: \vec{\vartheta } = \lambda \vec{c}\times \left ( \vec{a}\times \vec{b} \right )$
(a vector in the plane of $\vec{a}$ & $\vec{b}$ and perpendicular  to $\vec{c}$)
$\vec{c}\times \left ( \vec{a}\times \vec{b} \right )= 14\hat{i}-12\hat{j}+18\hat{k}$
$\vec{\vartheta }= \lambda\left ( 14\hat{i}-12\hat{j}+18\hat{k} \right )$
$Also\,\; \vec{\vartheta }\cdot \hat{a}= 19$
$\Rightarrow \lambda \left ( 14\hat{i}-12\hat{j}+18\hat{k} \right )\cdot \frac{\left ( 2\hat{i} -\hat{j}+2\hat{k}\right )}{\sqrt{2^{2}+1^{1}+2^{2}}}= 19$
$\Rightarrow \lambda = \frac{19\times 3}{28+12+36}= \frac{19\times 3}{76}= \frac{3}{4}$
$so\: \left | 2\vartheta ^{2} \right |= \left | 2\times \frac{3}{4}\left ( 14\hat{i}-12\hat{j}+18\hat{k} \right ) \right |^{2}$
$= 9\left ( 7^{2}+6^{2}+9^{2} \right )$
$= 1494$