Given triangle $ABC$ with vertices $A(2, 6)$, $B(2+2\sqrt{2}, 0, 4)$, and $C(2+2\sqrt{2}, 4, 4)$. We need to find the lengths of sides $AB$ and $AC$, and the measure of angle $\angle BAC$. Also, identify the type of triangle $ABC$.

1. Problem Description

Given triangle $ABC$ with vertices $A(2, 6)$, $B(2+2\sqrt{2}, 0, 4)$, and $C(2+2\sqrt{2}, 4, 4)$. We need to find the lengths of sides $AB$ and $AC$, and the measure of angle $\angle BAC$. Also, identify the type of triangle $ABC$.

2. Solution Steps

First, calculate the lengths of the sides $AB$ and $AC$. The distance between two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ in 3D space is given by:

$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$

The length of side $AB$ is:

$AB = \sqrt{((2+2\sqrt{2}) - 2)^2 + (0 - 6)^2 + (4 - 6)^2} = \sqrt{(2\sqrt{2})^2 + (-6)^2 + (-2)^2} = \sqrt{8 + 36 + 4} = \sqrt{48} = 4\sqrt{3}$

The length of side $AC$ is:

$AC = \sqrt{((2+2\sqrt{2}) - 2)^2 + (4 - 6)^2 + (4 - 6)^2} = \sqrt{(2\sqrt{2})^2 + (-2)^2 + (-2)^2} = \sqrt{8 + 4 + 4} = \sqrt{16} = 4$

Next, calculate the length of side $BC$:

$BC = \sqrt{((2+2\sqrt{2}) - (2+2\sqrt{2}))^2 + (4 - 0)^2 + (4 - 4)^2} = \sqrt{(0)^2 + (4)^2 + (0)^2} = \sqrt{16} = 4$

Since $AC = BC = 4$, the triangle is an isosceles triangle.

Now, we need to find the angle $\angle BAC$. We can use the Law of Cosines:

$BC^2 = AB^2 + AC^2 - 2 \cdot AB \cdot AC \cdot \cos(\angle BAC)$

$4^2 = (4\sqrt{3})^2 + 4^2 - 2 \cdot 4\sqrt{3} \cdot 4 \cdot \cos(\angle BAC)$

$16 = 48 + 16 - 32\sqrt{3} \cdot \cos(\angle BAC)$

$16 = 64 - 32\sqrt{3} \cdot \cos(\angle BAC)$

$-48 = -32\sqrt{3} \cdot \cos(\angle BAC)$

$\cos(\angle BAC) = \frac{-48}{-32\sqrt{3}} = \frac{3}{2\sqrt{3}} = \frac{3\sqrt{3}}{2 \cdot 3} = \frac{\sqrt{3}}{2}$

$\angle BAC = \arccos(\frac{\sqrt{3}}{2}) = 30^\circ$

Since $\angle BAC = 30^\circ$ and $AC = BC$, the other two angles are equal: $\angle ABC = \angle BAC = (180^\circ - 30^\circ)/2 = 75^\circ$.

3. Final Answer

$AB = 4\sqrt{3}$

$AC = 4$

$\angle BAC = 30^\circ$

The triangle $ABC$ is an isosceles triangle.