Let $$a = 2i + j - 2k$$    and $$b = i + j.$$   If $$c$$ is a vector such that $$a.\,\,c = \left| c \right|,\,\,\left| {c - a} \right| = 2\sqrt 2 $$      and the angle between $$\left( {a \times b} \right)$$  and $$c$$ is $${30^ \circ },$$  then $$\left| {\left( {a \times b} \right) \times c} \right| = ?$$

Solution :
$$\eqalign{ & \left| {\left( {\vec a \times \vec b} \right) \times \vec c} \right| = \left| {\vec a \times \vec b} \right|\left| {\vec c} \right|\,\sin \,{30^ \circ } \cr & = \frac{1}{2}\left| {\vec a \times \vec b} \right|\left| {\vec c} \right|.....(1) \cr & {\text{we have, }}\vec a = 2\hat i + \hat j - 2\hat k{\text{ and }}\vec b = \hat i + \hat j \cr & \Rightarrow \vec a \times \vec b = 2\hat i - 2\hat j + \hat k\,\,\,\,\, \Rightarrow \left| {\vec a \times \vec b} \right| = \sqrt 9 = 3 \cr & {\text{Also given }}\left| {\vec c - \vec a} \right| = 2\sqrt 2 \cr & \Rightarrow {\left| {\vec c - \vec a} \right|^2} = 8 \cr & \Rightarrow \left( {\vec c - \vec a} \right).\left( {\vec c - \vec a} \right) = 8 \cr & \Rightarrow {\left| {\vec c} \right|^2} - \vec c.\vec a - \vec a.\vec c + {\left| {\vec a} \right|^2} = 8 \cr & {\text{As }}\left| {\vec a} \right| = 3{\text{ and }}\vec a.\vec c = \left| {\vec c} \right|,{\text{we get}} \cr & \Rightarrow {\left| {\vec c} \right|^2} - 2\left| {\vec c} \right| + 1 = 0 \cr & \Rightarrow \,{\left( {\left| {\vec c} \right| - 1} \right)^2} = 0 \cr & \Rightarrow \left| {\vec c} \right| = 1 \cr} $$
Substituting values of $$\left| {\vec a \times \vec b} \right|$$  and $$\left| {\vec c} \right|$$ in (1), we get
$$\left| {\left( {\vec a \times \vec b} \right) \times \vec c} \right| = \frac{1}{2} \times 3 \times 1 = \frac{3}{2}$$