If the position vectors of three points A, B, C are respectively , and , find the unit vector…

Let the position vectors of the vertices be $\vec{a} = \hat{i} + \hat{j} + \hat{k}$, $\vec{b} = 2\hat{i} + 3\hat{j} - 4\hat{k}$, $\vec{c} = 7\hat{i} + 4\hat{j} + 9\hat{k}$. $\vec{AB} = \vec{b} - \vec{a} = (2-1)\hat{i} + (3-1)\hat{j} + (-4-1)\hat{k} = \hat{i} + 2\hat{j} - 5\hat{k}$. $\vec{AC} = \vec{c} - \vec{a} = (7-1)\hat{i} + (4-1)\hat{j} + (9-1)\hat{k} = 6\hat{i} + 3\hat{j} + 8\hat{k}$. A vector perpendicular to the plane of triangle ABC is $\vec{n} = \vec{AB} \times \vec{AC}$: $\vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & -5 \\ 6 & 3 & 8 \end{vmatrix} = \hat{i}(16 + 15) - \hat{j}(8 + 30) + \hat{k}(3 - 12) = 31\hat{i} - 38\hat{j} - 9\hat{k}$. Magnitude: $|\vec{n}| = \sqrt{31^2 + (-38)^2 + (-9)^2} = \sqrt{961 + 1444 + 81} = \sqrt{2486}$. Therefore, the unit vector perpendicular to the plane is: $\hat{n} = \pm \frac{31\hat{i} - 38\hat{j} - 9\hat{k}}{\sqrt{2486}}$.