Find the area of a parallelogram formechy the vectors \( \overrightarrow{P Q} \) and \( \overrightarrow{P R} \), where \( P=(0,2,-6) \), \( Q=(6,-2,-6) \) and \( R=(4,-1,5) \).

To find the area of the parallelogram formed by the vectors \( \overrightarrow{P Q} \) and \( \overrightarrow{P R} \), we first need to determine these vectors. 1. Calculate \( \overrightarrow{P Q} \): \[ \overrightarrow{P Q} = Q - P = (6, -2, -6) - (0, 2, -6) = (6 - 0, -2 - 2, -6 - (-6)) = (6, -4, 0) \] 2. Calculate \( \overrightarrow{P R} \): \[ \overrightarrow{P R} = R - P = (4, -1, 5) - (0, 2, -6) = (4 - 0, -1 - 2, 5 - (-6)) = (4, -3, 11) \] Next, we find the area of the parallelogram using the magnitude of the cross product of \( \overrightarrow{P Q} \) and \( \overrightarrow{P R} \). 3. Calculate the cross product \( \overrightarrow{P Q} \times \overrightarrow{P R} \): \[ \overrightarrow{P Q} \times \overrightarrow{P R} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 6 & -4 & 0 \\ 4 & -3 & 11 \end{vmatrix} \] Calculating the determinant: \[ = \mathbf{i} \begin{vmatrix}-4 & 0 \\ -3 & 11\end{vmatrix} - \mathbf{j} \begin{vmatrix}6 & 0 \\ 4 & 11\end{vmatrix} + \mathbf{k} \begin{vmatrix}6 & -4 \\ 4 & -3\end{vmatrix} \] \[ = \mathbf{i}((-4)(11) - (0)(-3)) - \mathbf{j}((6)(11) - (0)(4)) + \mathbf{k}((6)(-3) - (-4)(4)) \] \[ = \mathbf{i}(-44) - \mathbf{j}(66) + \mathbf{k}(-18 + 16) \] \[ = -44\mathbf{i} - 66\mathbf{j} - 2\mathbf{k} \] So, the cross product is: \[ \overrightarrow{P Q} \times \overrightarrow{P R} = (-44, -66, -2) \] 4. Calculate the magnitude of the cross product: \[ \left| \overrightarrow{P Q} \times \overrightarrow{P R} \right| = \sqrt{(-44)^2 + (-66)^2 + (-2)^2} \] \[ = \sqrt{1936 + 4356 + 4} = \sqrt{6296} \] 5. Now simplify \( \sqrt{6296} \): \[ 6296 = 4 \times 1574 = 4 \times 2 \times 787 = 8 \times 787 \] Thus, \[ \left| \overrightarrow{P Q} \times \overrightarrow{P R} \right| = 2\sqrt{1574} \] Finally, the area of the parallelogram is: \[ \text{Area} = \left| \overrightarrow{P Q} \times \overrightarrow{P R} \right| = 2\sqrt{1574} \]