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(A) Given $\overline{OP} = -\hat{i}-2\hat{j}+3\hat{k}$.$\overline{AB} = \overline{OB} - \overline{OA} = (2\hat{i}+4\hat{j}-2\hat{k}) - (-2\hat{i}+\hat

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$3$

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(A) Given $\overline{OP} = -\hat{i}-2\hat{j}+3\hat{k}$.$\overline{AB} = \overline{OB} - \overline{OA} = (2\hat{i}+4\hat{j}-2\hat{k}) - (-2\hat{i}+\hat{j}-3\hat{k}) = 4\hat{i}+3\hat{j}+\hat{k}$.$\overline{AC} = \overline{OC} - \overline{OA} = (-4\hat{i}+2\hat{j}-\hat{k}) - (-2\hat{i}+\hat{j}-3\hat{k}) = -2\hat{i}+\hat{j}+2\hat{k}$.$A$ vector perpendicular to both $\overline{AB}$ and $\overline{AC}$ is $\vec{n} = \overline{AB} 	imes \overline{AC}$.$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 4 & 3 & 1 \ -2 & 1 & 2 \end{vmatrix} = \hat{i}(6-1) - \hat{j}(8+2) + \hat{k}(4+6) = 5\hat{i}-10\hat{j}+10\hat{k}$.The projection of $\overline{OP}$ on $\vec{n}$ is given by $\frac{|\overline{OP} \cdot \vec{n}|}{|\vec{n}|}$.$\overline{OP} \cdot \vec{n} = (-1)(5) + (-2)(-10) + (3)(10) = -5 + 20 + 30 = 45$.$|\vec{n}| = \sqrt{5^2 + (-10)^2 + 10^2} = \sqrt{25 + 100 + 100} = \sqrt{225} = 15$.Projection $= \frac{|45|}{15} = 3$.

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