1. Key Concepts and Formulas

• Area of a Quadrilateral using Diagonals: The area of a quadrilateral can be computed as half the magnitude of the cross product of its diagonal vectors. If $\vec{d_1}$ and $\vec{d_2}$ are the vectors representing the diagonals, the area is given by $\text{Area} = \frac{1}{2} | \vec{d_1} \times \vec{d_2} |$.
• Vector Between Two Points: The vector from point $P_1(x_1, y_1, z_1)$ to $P_2(x_2, y_2, z_2)$ is calculated as $\overrightarrow{P_1P_2} = (x_2-x_1)\hat{i} + (y_2-y_1)\hat{j} + (z_2-z_1)\hat{k}$.
• Cross Product of Vectors: For two vectors $\vec{a} = a_x\hat{i} + a_y\hat{j} + a_z\hat{k}$ and $\vec{b} = b_x\hat{i} + b_y\hat{j} + b_z\hat{k}$, their cross product is given by: $\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a_x & a_y & a_z \\ b_x & b_y & b_z \end{vmatrix}$
• Magnitude of a Vector: The magnitude of a vector $\vec{v} = v_x\hat{i} + v_y\hat{j} + v_z\hat{k}$ is $| \vec{v} | = \sqrt{v_x^2 + v_y^2 + v_z^2}$.

2. Step-by-Step Solution

We are given the vertices of the quadrilateral ABCD as A$(2,1,1)$, B$(1,2,5)$, C$(-2,-3,5)$, and D$(1,-6,-7)$.

Step 1: Calculate the vectors representing the diagonals. To find the area using the diagonal method, we first need to determine the vectors corresponding to the diagonals $\overrightarrow{\mathrm{AC}}$ and $\overrightarrow{\mathrm{BD}}$.

• Diagonal $\overrightarrow{\mathrm{AC}}$: Using points A$(2,1,1)$ and C$(-2,-3,5)$: $\overrightarrow{\mathrm{AC}} = (-2 - 2)\hat{i} + (-3 - 1)\hat{j} + (5 - 1)\hat{k}$ $\overrightarrow{\mathrm{AC}} = -4\hat{i} - 4\hat{j} + 4\hat{k}$

• Diagonal $\overrightarrow{\mathrm{BD}}$: Using points B$(1,2,5)$ and D$(1,-6,-7)$: $\overrightarrow{\mathrm{BD}} = (1 - 1)\hat{i} + (-6 - 2)\hat{j} + (-7 - 5)\hat{k}$ $\overrightarrow{\mathrm{BD}} = 0\hat{i} - 8\hat{j} - 12\hat{k}$ $\overrightarrow{\mathrm{BD}} = -8\hat{j} - 12\hat{k}$

Step 2: Calculate the cross product of the diagonal vectors. Now, we compute the cross product $\overrightarrow{\mathrm{AC}} \times \overrightarrow{\mathrm{BD}}$ using the determinant formula: $\overrightarrow{\mathrm{AC}} \times \overrightarrow{\mathrm{BD}} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -4 & -4 & 4 \\ 0 & -8 & -12 \end{vmatrix}$ Expanding the determinant: $= \hat{i} \left( (-4)(-12) - (4)(-8) \right) - \hat{j} \left( (-4)(-12) - (4)(0) \right) + \hat{k} \left( (-4)(-8) - (-4)(0) \right)$ $= \hat{i} (48 - (-32)) - \hat{j} (48 - 0) + \hat{k} (32 - 0)$ $= \hat{i} (48 + 32) - 48\hat{j} + 32\hat{k}$ $= 80\hat{i} - 48\hat{j} + 32\hat{k}$

Step 3: Calculate the magnitude of the cross product. Next, we find the magnitude of the resulting cross product vector, $80\hat{i} - 48\hat{j} + 32\hat{k}$. $| \overrightarrow{\mathrm{AC}} \times \overrightarrow{\mathrm{BD}} | = \sqrt{(80)^2 + (-48)^2 + (32)^2}$ To simplify the calculation, we can factor out a common term. Notice that $80 = 16 \times 5$, $48 = 16 \times 3$, and $32 = 16 \times 2$. So, $80\hat{i} - 48\hat{j} + 32\hat{k} = 16(5\hat{i} - 3\hat{j} + 2\hat{k})$. The magnitude becomes: $| \overrightarrow{\mathrm{AC}} \times \overrightarrow{\mathrm{BD}} | = | 16(5\hat{i} - 3\hat{j} + 2\hat{k}) |$ $= 16 | 5\hat{i} - 3\hat{j} + 2\hat{k} |$ $= 16 \sqrt{(5)^2 + (-3)^2 + (2)^2}$ $= 16 \sqrt{25 + 9 + 4}$ $= 16 \sqrt{38}$

Step 4: Apply the area formula. Finally, we use the formula for the area of a quadrilateral: $\text{Area} = \frac{1}{2} | \overrightarrow{\mathrm{AC}} \times \overrightarrow{\mathrm{BD}} |$ $\text{Area} = \frac{1}{2} (16 \sqrt{38})$ $\text{Area} = 8 \sqrt{38} \text{ square units}$

3. Common Mistakes & Tips

• Vector Direction: Ensure the vector subtraction is done correctly ($P_2 - P_1$) to maintain the correct direction of the diagonal vectors.
• Cross Product Signs: Be meticulous with the signs when calculating the components of the cross product, especially for the $\hat{j}$ term.
• Simplifying Radicals: Factoring out common terms before squaring can significantly simplify the magnitude calculation and prevent arithmetic errors.

4. Summary The area of a quadrilateral can be efficiently calculated using the magnitudes of the cross product of its diagonal vectors. By computing the vectors $\overrightarrow{\mathrm{AC}}$ and $\overrightarrow{\mathrm{BD}}$, finding their cross product, and then determining the magnitude of this cross product, we arrived at the area of the quadrilateral ABCD as $8 \sqrt{38}$ square units.

The final answer is $\boxed{8 \sqrt{38}}$.