Question:

To find the area of the quadrilateral $ABCD$, we can use the formula for the area of a quadrilateral given its vertices. This involves calculating the cross product of vectors formed from the vertices. Step 1: Find the vectors $\overrightarrow{AC}$ and $\overrightarrow{BD}$. Given $A(1,-1,2)$, $B(5,7,-6)$, $C(3,4,-10)$, and $D(-1,-4,-2)$, we can find $\overrightarrow{AC}$ and $\overrightarrow{BD}$ as follows: $\overrightarrow{AC} = C - A = (3 - 1, 4 - (-1), -10 - 2) = (2, 5, -12)$ $\overrightarrow{BD} = D - B = (-1 - 5, -4 - 7, -2 - (-6)) = (-6, -11, 4)$ Step 2: Calculate the cross product $\overrightarrow{AC} \times \overrightarrow{BD}$. The cross product of two vectors $\mathbf{u} = (u_1, u_2, u_3)$ and $\mathbf{v} = (v_1, v_2, v_3)$ is given by $$ \begin{aligned} \mathbf{u} \times \mathbf{v} &= \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ u_1 & u_2 & u_3 \\ v_1 & v_2 & v_3 \\ \end{vmatrix} \\ &= (u_2v_3 - u_3v_2)\mathbf{i} - (u_1v_3 - u_3v_1)\mathbf{j} + (u_1v_2 - u_2v_1)\mathbf{k} \end{aligned} $$ Applying this formula to $\overrightarrow{AC}$ and $\overrightarrow{BD}$: $$ \begin{aligned} \overrightarrow{AC} \times \overrightarrow{BD} &= \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 5 & -12 \\ -6 & -11 & 4 \\ \end{vmatrix} \\ &= (5 \cdot 4 - (-12) \cdot (-11))\mathbf{i} - (2 \cdot 4 - (-12) \cdot (-6))\mathbf{j} + (2 \cdot (-11) - 5 \cdot (-6))\mathbf{k} \\ &= (20 - 132)\mathbf{i} - (8 - 72)\mathbf{j} + (-22 + 30)\mathbf{k} \\ &= -112\mathbf{i} + 64\mathbf{j} + 8\mathbf{k} \\ &= (-112, 64, 8) \end{aligned} $$ Step 3: Calculate the magnitude of $\overrightarrow{AC} \times \overrightarrow{BD}$. The magnitude of a vector $\mathbf{v} = (v_1, v_2, v_3)$ is given by $|\mathbf{v}| = \sqrt{v_1^2 + v_2^2 + v_3^2}$. Thus, the magnitude of $\overrightarrow{AC} \times \overrightarrow{BD}$ is $$ \begin{aligned} |\overrightarrow{AC} \times \overrightarrow{BD}| &= \sqrt{(-112)^2 + 64^2 + 8^2} \\ &= \sqrt{12544 + 4096 + 64} \\ &= \sqrt{16704} \\ &= \sqrt{16 \cdot 1044} \\ &= 4\sqrt{1044} \\ &= 4\sqrt{4 \cdot 261} \\ &= 4 \cdot 2\sqrt{261} \\ &= 8\sqrt{261} \\ &= 8\sqrt{3 \cdot 87} \\ &= 8\sqrt{3 \cdot 3 \cdot 29} \\ &= 8 \cdot 3\sqrt{29} \\ &= 24\sqrt{29} \end{aligned} $$ Step 4: Calculate the area of the quadrilateral $ABCD$. The area of the quadrilateral can be found using the formula $\text{Area} = \frac{1}{2}|\overrightarrow{AC} \times \overrightarrow{BD}|$. Substituting the magnitude calculated in Step 3: $$ \begin{aligned} \text{Area} &= \frac{1}{2} \cdot 24\sqrt{29} \\ &= 12\sqrt{29} \end{aligned} $$ Therefore: $12\sqrt{29}$, which corresponds to option 2. <div class="key-concept"><strong>Key Concept:</strong> Area $=\frac{1}{2}|\overrightarrow{AC}\times\overrightarrow{BD}|$. $\overrightarrow{AC}=(2,5,-12)$, $\overrightarrow{BD}=(6,11,-4)$. Cross product $=(2,5,-12)\times(6,11,-4)$.</div> <div class="trap-box"><strong>Trap:</strong> $\overrightarrow{AC}\times\overrightarrow{BD}=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\2&5&-12\\6&11&-4\end{vmatrix}=(112,-64,-8)=8(14,-8,-1)$. $|\overrightarrow{AC}\times\overrightarrow{BD}|=8\sqrt{196+64+1}=8\sqrt{261}=8\cdot3\sqrt{29}=24\sqrt{29}$. Area $=\frac{1}{2}\cdot24\sqrt{29}=12\sqrt{29}$.</div>