${\Rightarrow|\vec{a}|=3}$
and $\vec a \times \vec b = \begin{array}{*{20}{c}}
{\hat i}&{\hat j}&{\hat k}\\
2&1&{ - 2}\\
1&1&0
\end{array} = 2\hat i - 2\hat j + \hat k$
$|\vec{a} \times \vec{b}|=\sqrt{4}+4+1=3$
$\text { Now, }|\vec{c}-\vec{a}|=2 \sqrt{2} \Rightarrow|\vec{c}-\vec{a}|^{2}=8$
$\Rightarrow|\vec{c}-\vec{a}| \cdot(\vec{c}-\vec{a})=8$
${\Rightarrow|\vec{c}|^{2}+|\vec{a}|^{2}-2 \vec{c} \cdot \vec{a}=8}$
${\Rightarrow|\vec{c}|^{2}+9-2|\vec{c}|=8} $
$ \Rightarrow {(\left| {\vec c} \right| - 1)^2} = 0 \Rightarrow |\vec c| = 1$
$\therefore|(\vec{a} \times \vec{b}) \times \vec{c}|=|\vec{a} \times \vec{b}||\vec{c}| \sin 30^{\circ}$
$=3 \times 1 \times \frac{1}{2}=\frac{3}{2}$
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$\left( {1 + \alpha } \right)x + \beta y + z = 2$ ; $\alpha x + \left( {1 + \beta } \right)y + z = 3$ ; $\alpha x + \beta y + 2z = 2$ has a unique solution, is