Let $$A(2, 3, 5)$$ and $$C(-3, 4, -2)$$ be opposite vertices of a parallelogram $$ABCD$$ if the diagonal $$\vec{BD} = \hat{i} + 2\hat{j} + 3\hat{k}$$ then the area of the parallelogram is equal to

Finding Diagonal $$\vec{AC}$$

$$\vec{AC} = (-3-2)\hat{i} + (4-3)\hat{j} + (-2-5)\hat{k} = -5\hat{i} + \hat{j} - 7\hat{k}$$

Area $$= \frac{1}{2} |\vec{AC} \times \vec{BD}|$$

$$\vec{AC} \times \vec{BD} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -5 & 1 & -7 \\ 1 & 2 & 3 \end{vmatrix} = \hat{i}(3+14) - \hat{j}(-15+7) + \hat{k}(-10-1) = 17\hat{i} + 8\hat{j} - 11\hat{k}$$

• Magnitude $$= \sqrt{17^2 + 8^2 + (-11)^2} = \sqrt{289 + 64 + 121} = \sqrt{474}$$

• Area $$= \mathbf{\frac{1}{2}\sqrt{474}}$$ (Option B)