Find the area of the triangle ABC, given the coordinates of the vertices A(2, 2, 6), B(2 + $2\sqrt{2}$, 0, 4), and C(2 + $2\sqrt{2}$, 4, 4).

1. Problem Description

Find the area of the triangle ABC, given the coordinates of the vertices A(2, 2, 6), B(2 + $2\sqrt{2}$, 0, 4), and C(2 + $2\sqrt{2}$, 4, 4).

2. Solution Steps

First, we find the vectors $\vec{AB}$ and $\vec{AC}$.

$\vec{AB} = B - A = (2 + 2\sqrt{2} - 2, 0 - 2, 4 - 6) = (2\sqrt{2}, -2, -2)$

$\vec{AC} = C - A = (2 + 2\sqrt{2} - 2, 4 - 2, 4 - 6) = (2\sqrt{2}, 2, -2)$

Next, we find the cross product of these two vectors.

$\vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2\sqrt{2} & -2 & -2 \\ 2\sqrt{2} & 2 & -2 \end{vmatrix} = \hat{i}((-2)(-2) - (-2)(2)) - \hat{j}((2\sqrt{2})(-2) - (-2)(2\sqrt{2})) + \hat{k}((2\sqrt{2})(2) - (-2)(2\sqrt{2}))$

$= \hat{i}(4 + 4) - \hat{j}(-4\sqrt{2} + 4\sqrt{2}) + \hat{k}(4\sqrt{2} + 4\sqrt{2}) = 8\hat{i} + 0\hat{j} + 8\sqrt{2}\hat{k} = (8, 0, 8\sqrt{2})$

The magnitude of the cross product is

$||\vec{AB} \times \vec{AC}|| = \sqrt{8^2 + 0^2 + (8\sqrt{2})^2} = \sqrt{64 + 0 + 64 \cdot 2} = \sqrt{64 + 128} = \sqrt{192} = \sqrt{64 \cdot 3} = 8\sqrt{3}$

The area of the triangle ABC is half the magnitude of the cross product.

$Area = \frac{1}{2} ||\vec{AB} \times \vec{AC}|| = \frac{1}{2}(8\sqrt{3}) = 4\sqrt{3}$

3. Final Answer

The area of the triangle ABC is $4\sqrt{3}$.