nd the area of a triangle \( P Q R \), where \( P=(-4,0,-6), Q=(0,-3,5) \), and \( R=(5,4,5) \)

To find the area of triangle \( PQR \) with vertices \( P=(-4,0,-6) \), \( Q=(0,-3,5) \), and \( R=(5,4,5) \), we can use the formula for the area based on the cross product of vectors. First, we need to find the vectors \( \overrightarrow{PQ} \) and \( \overrightarrow{PR} \): \[ \overrightarrow{PQ} = Q - P = (0 - (-4), -3 - 0, 5 - (-6)) = (4, -3, 11) \] \[ \overrightarrow{PR} = R - P = (5 - (-4), 4 - 0, 5 - (-6)) = (9, 4, 11) \] Next, we'll compute the cross product \( \overrightarrow{PQ} \times \overrightarrow{PR} \): \[ \overrightarrow{PQ} \times \overrightarrow{PR} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 4 & -3 & 11 \\ 9 & 4 & 11 \end{vmatrix} \] Calculating this determinant gives: \[ \mathbf{i}((-3)(11) - (11)(4)) - \mathbf{j}((4)(11) - (11)(9)) + \mathbf{k}((4)(4) - (-3)(9)) \] \[ = \mathbf{i}(-33 - 44) - \mathbf{j}(44 - 99) + \mathbf{k}(16 + 27) \] \[ = \mathbf{i}(-77) - \mathbf{j}(-55) + \mathbf{k}(43) \] \[ = (-77, 55, 43) \] Now, to find the magnitude of the cross product: \[ \|\overrightarrow{PQ} \times \overrightarrow{PR}\| = \sqrt{(-77)^2 + 55^2 + 43^2} \] Calculating the squares: \[ = \sqrt{5929 + 3025 + 1849} = \sqrt{10803} \] Finally, the area \( A \) of triangle \( PQR \) is half the magnitude of the cross product: \[ A = \frac{1}{2} \|\overrightarrow{PQ} \times \overrightarrow{PR}\| = \frac{1}{2} \sqrt{10803} \] Thus, the area of triangle \( PQR \) is: \[ A = \frac{\sqrt{10803}}{2} \] You can simplify \( \sqrt{10803} \), which is approximately \( 103.9 \). Therefore, the area is roughly \( 52.0 \) square units.