\qquad \vec{b}=\hat{i}-\hat{j}+\hat{k} $$

So,

$$ \vec{a}\times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ 1 & 2 & -2\\ 1 & -1 & 1 \end{vmatrix} $$

$$ \vec{a}\times \vec{b} =0\hat{i}-3\hat{j}-3\hat{k} $$

Hence,

$$ |\vec{a}\times \vec{b}|=\sqrt{0^2+(-3)^2+(-3)^2} =\sqrt{18}=3\sqrt{2} $$

Now use the given conditions involving $\vec{c}$.

We have

$$ \vec{a}\cdot \vec{c}=|\vec{c}| $$

Also,

$$ |\vec{c}-\vec{a}|=2\sqrt{2} $$

Squaring both sides,

$$ |\vec{c}-\vec{a}|^2=8 $$

Using the formula

$$ |\vec{c}-\vec{a}|^2=|\vec{c}|^2+|\vec{a}|^2-2\vec{a}\cdot \vec{c} $$

Now,

$$ |\vec{a}|=\sqrt{1^2+2^2+(-2)^2}=3 $$

So,

$$ |\vec{c}|^2+9-2|\vec{c}|=8 $$

$$ |\vec{c}|^2-2|\vec{c}|+1=0 $$

$$ (|\vec{c}|-1)^2=0 $$

Thus,

$$ |\vec{c}|=1 $$

The angle between $\vec{a}\times \vec{b}$ and $\vec{c}$ is $60^\circ$.

So,

$$ |(\vec{a}\times \vec{b})\times \vec{c}| = |\vec{a}\times \vec{b}|\,|\vec{c}|\sin 60^\circ $$

$$ =3\sqrt{2}\times 1 \times \frac{\sqrt{3}}{2} $$

$$ =\frac{3\sqrt{6}}{2} $$

Since

$$ \frac{3\sqrt{6}}{2}=3\sqrt{\frac{3}{2}} $$

the correct option is

$$ \boxed{3\sqrt{\frac{3}{2}}} $$

Option C.

Vector Algebra

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