To find the area of the quadrilateral ABCD, given the vectors representing its diagonals, \(\overrightarrow{AC}\) and \(\overrightarrow{BD}\), we can use the property that the area is half the magnitude of the cross product of the diagonals.

Finding the Diagonals

The diagonals are given as:

• \(\overrightarrow{AC} = 2\hat{i} + \hat{j} + \hat{k}\)
• \(\overrightarrow{BD} = -\hat{i} + 3\hat{j} + 2\hat{k}\)

Calculating the Cross Product of Diagonals

The cross product \(\overrightarrow{AC} \times \overrightarrow{BD}\) is calculated as follows:

\(\overrightarrow{AC} \times \overrightarrow{BD} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & 1 \\ -1 & 3 & 2 \end{vmatrix}\)

Expanding the determinant:

\(= \hat{i}((1)(2) - (1)(3)) - \hat{j}((2)(2) - (1)(-1)) + \hat{k}((2)(3) - (1)(-1))\)

\(= \hat{i}(2 - 3) - \hat{j}(4 - (-1)) + \hat{k}(6 - (-1))\)

\(= \hat{i}(-1) - \hat{j}(4 + 1) + \hat{k}(6 + 1)\)

\(= -1\hat{i} - 5\hat{j} + 7\hat{k}\)

Determining the Magnitude of the Cross Product

Next, we find the magnitude of the resulting vector:

\(|\overrightarrow{AC} \times \overrightarrow{BD}| = |-1\hat{i} - 5\hat{j} + 7\hat{k}|\)

\(= \sqrt{(-1)^2 + (-5)^2 + (7)^2}\)

\(= \sqrt{1 + 25 + 49}\)

\(= \sqrt{75}\)

\(= \sqrt{25 \times 3}\)

\(= 5\sqrt{3}\)

Calculating the Area of the Quadrilateral

The area of the quadrilateral ABCD is half the magnitude of the cross product of its diagonals:

Area = \(\dfrac{1}{2} |\overrightarrow{AC} \times \overrightarrow{BD}|\)

Area = \(\dfrac{1}{2} (5\sqrt{3})\)

Area = \(\dfrac{5}{2}\sqrt{3}\)

Conclusion

The area of the quadrilateral ABCD is \(\dfrac{5}{2}\sqrt{3}\).