Let $$\vec{a} = 6\hat{i} + \hat{j} - \hat{k}$$ and $$\vec{b} = \hat{i} + \hat{j}$$. If $$\vec{c}$$ is a vector such that $$|\vec{c}| \geq 6$$, $$\vec{a} \cdot \vec{c} = 6|\vec{c}|$$, $$|\vec{c} - \vec{a}| = 2\sqrt{2}$$ and the angle between $$\vec{a} \times \vec{b}$$ and $$\vec{c}$$ is $$60°$$, then $$|(\vec{a} \times \vec{b}) \times \vec{c}|$$ is equal to:

|a| = √(36+1+1) = √38. a·c = 6|c|: cos θ_ac = 6/√38.

|c-a|² = |c|²-2a·c+|a|² = |c|²-12|c|+38 = 8. So |c|²-12|c|+30 = 0. |c| = (12±√(144-120))/2 = (12±√24)/2 = 6±√6. Since |c|≥6: |c| = 6+√6.

a×b = |i j k; 6 1 -1; 1 1 0| = (0+1)i-(-1+0... = i-(-1)j+(6-1)k... Let me compute: = (0+1)i-(0+1)j+(6-1)k = i-j+5k. |a×b| = √(1+1+25) = √27 = 3√3.

|(a×b)×c| = |a×b|·|c|·sin60° = 3√3·(6+√6)·(√3/2) = (9/2)(6+√6).

The correct answer is Option (3): 9(6+√6)/2.