Let $$\vec a = 2\hat i + \hat j - 2\hat k$$    and $$\vec b = \hat i + \hat j.$$   Let $${\vec c}$$ be a vector such that $$\left| {\vec c - \vec a} \right| = 3,\,\left| {\left( {\vec a \times \vec b} \right) \times \vec c} \right| = 3$$      and the angle between $${\vec c}$$ and $$\vec a \times \vec b$$  be $${30^ \circ }.$$  Then $$\vec a.\vec c$$  is equal to :

Solution :
Given :
$$\eqalign{ & \vec a = 2\hat i + \hat j - 2\hat k,\,\vec b = \hat i + \hat j \cr & \Rightarrow \left| {\vec a} \right| = 3 \cr & \therefore \,\vec a \times \vec b = 2\hat i - 2\hat j + \hat k \cr & \left| {\vec a \times \vec b} \right| = \sqrt {{2^2} + {2^2} + {1^2}} = 3 \cr} $$
We have $$\left| {\left( {\vec a \times \vec b} \right) \times \vec c} \right| = \left| {\vec a \times \vec b} \right|\left| {\vec c} \right|\,\sin \,{30^ \circ }$$
$$\eqalign{ & \Rightarrow \left| {\left( {\vec a \times \vec b} \right) \times \vec c} \right| = 3\left| {\vec c} \right|.\frac{1}{2} \cr & \Rightarrow 3 = 3\left| {\vec c} \right|.\frac{1}{2} \cr & \therefore \left| {\vec c} \right| = 2 \cr} $$
Now $$\left| {\vec c - \vec a} \right| = 3$$
On squaring, we get
$$\eqalign{ & \Rightarrow {c^2} + {a^2} - 2\vec c.\vec a = 9 \cr & \Rightarrow 4 + 9 - 2\vec a.\vec c = 9 \cr & \Rightarrow \vec a.\vec c = 2\,\,\,\,\,\,\,\,\,\,\left[ {\because \vec c.\vec a = \vec a.\vec c} \right] \cr} $$