Let $$O$$ be the origin and the position vector of the point $$P$$ be $$-\hat{i} - 2\hat{j} + 3\hat{k}$$. If the position vectors of the points $$A$$, $$B$$ and $$C$$ are $$-2\hat{i} + \hat{j} - 3\hat{k}$$, $$2\hat{i} + 4\hat{j} - 2\hat{k}$$ and $$-4\hat{i} + 2\hat{j} - \hat{k}$$ respectively, then the projection of the vector $$\overrightarrow{OP}$$ on a vector perpendicular to the vectors $$\overrightarrow{AB}$$ and $$\overrightarrow{AC}$$ is

Let $$O$$ be the origin, with $$P = (-1, -2, 3)$$, $$A = (-2, 1, -3)$$, $$B = (2, 4, -2)$$, and $$C = (-4, 2, -1)$$. We seek the projection of $$\overrightarrow{OP}$$ onto a vector perpendicular to both $$\overrightarrow{AB}$$ and $$\overrightarrow{AC}$$.

We compute $$\overrightarrow{AB} = B - A = (2 - (-2), 4 - 1, -2 - (-3)) = (4, 3, 1)$$ and $$\overrightarrow{AC} = C - A = (-4 - (-2), 2 - 1, -1 - (-3)) = (-2, 1, 2)$$.

The cross product $$\vec{n} = \overrightarrow{AB} \times \overrightarrow{AC}$$ is given by the determinant $$\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & 3 & 1 \\ -2 & 1 & 2 \end{vmatrix} = \hat{i}(6 - 1) - \hat{j}(8 + 2) + \hat{k}(4 + 6) = 5\hat{i} - 10\hat{j} + 10\hat{k}$$, and its magnitude is $$|\vec{n}| = \sqrt{25 + 100 + 100} = \sqrt{225} = 15$$.

The vector $$\overrightarrow{OP}$$ is $$(-1, -2, 3)$$. Its projection onto $$\vec{n}$$ is $$\frac{\overrightarrow{OP} \cdot \vec{n}}{|\vec{n}|} = \frac{(-1)(5) + (-2)(-10) + (3)(10)}{15} = \frac{-5 + 20 + 30}{15} = \frac{45}{15} = 3$$.

The required projection equals 3.