The area of the quadrilateral $$ABCD$$ with vertices $$A(2, 1, 1)$$, $$B(1, 2, 5)$$, $$C(-2, -3, 5)$$ and $$D(1, -6, -7)$$ is equal to _____.

We need to find the area of quadrilateral $$ABCD$$ with vertices $$A(2,1,1)$$, $$B(1,2,5)$$, $$C(-2,-3,5)$$, and $$D(1,-6,-7)$$.

First, we check if the quadrilateral is planar.

Compute vectors from $$A$$:

$$\vec{AB} = B - A = (-1, 1, 4)$$

$$\vec{AC} = C - A = (-4, -4, 4)$$

$$\vec{AD} = D - A = (-1, -7, -8)$$

The scalar triple product is:

$$\vec{AB} \cdot (\vec{AC} \times \vec{AD})$$

First, $$\vec{AC} \times \vec{AD}$$:

$$\vec{AC} \times \vec{AD} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -4 & -4 & 4 \\ -1 & -7 & -8 \end{vmatrix}$$

$$= \hat{i}[(-4)(-8) - (4)(-7)] - \hat{j}[(-4)(-8) - (4)(-1)] + \hat{k}[(-4)(-7) - (-4)(-1)]$$

$$= \hat{i}[32 + 28] - \hat{j}[32 + 4] + \hat{k}[28 - 4]$$

$$= (60, -36, 24)$$

$$\vec{AB} \cdot (60, -36, 24) = (-1)(60) + (1)(-36) + (4)(24) = -60 - 36 + 96 = 0$$

Since the scalar triple product is $$0$$, the four points are coplanar.

Next, we split into two triangles and find total area.

We split $$ABCD$$ into triangles $$\triangle ABC$$ and $$\triangle ACD$$.

Area of $$\triangle ABC$$:

$$\vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 1 & 4 \\ -4 & -4 & 4 \end{vmatrix}$$

$$= \hat{i}[(1)(4) - (4)(-4)] - \hat{j}[(-1)(4) - (4)(-4)] + \hat{k}[(-1)(-4) - (1)(-4)]$$

$$= \hat{i}[4 + 16] - \hat{j}[-4 + 16] + \hat{k}[4 + 4]$$

$$= (20, -12, 8)$$

$$|\vec{AB} \times \vec{AC}| = \sqrt{400 + 144 + 64} = \sqrt{608} = 4\sqrt{38}$$

Area of $$\triangle ABC = \frac{1}{2} \times 4\sqrt{38} = 2\sqrt{38}$$

Area of $$\triangle ACD$$:

$$\vec{AC} \times \vec{AD} = (60, -36, 24)$$ (computed above)

$$|\vec{AC} \times \vec{AD}| = \sqrt{3600 + 1296 + 576} = \sqrt{5472} = 12\sqrt{38}$$

Area of $$\triangle ACD = \frac{1}{2} \times 12\sqrt{38} = 6\sqrt{38}$$

From this, we total area.

$$\text{Area of } ABCD = 2\sqrt{38} + 6\sqrt{38} = 8\sqrt{38}$$

The correct answer is Option B: $$8\sqrt{38}$$.