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(A) Let the position vectors of vertices $A, B,$ and $C$ be $\vec{a} = 3\hat{i}+4\hat{j}-\hat{k}$, $\vec{b} = \hat{i}+3\hat{j}+\hat{k}$, and $\vec{c}

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$\frac{4}{3}\sqrt{5}$

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(A) Let the position vectors of vertices $A, B,$ and $C$ be $\vec{a} = 3\hat{i}+4\hat{j}-\hat{k}$, $\vec{b} = \hat{i}+3\hat{j}+\hat{k}$, and $\vec{c} = 5\hat{i}+5\hat{j}+5\hat{k}$.The vectors representing the sides are:$\vec{AB} = \vec{b} - \vec{a} = (1-3)\hat{i} + (3-4)\hat{j} + (1-(-1))\hat{k} = -2\hat{i} - \hat{j} + 2\hat{k}$$\vec{AC} = \vec{c} - \vec{a} = (5-3)\hat{i} + (5-4)\hat{j} + (5-(-1))\hat{k} = 2\hat{i} + \hat{j} + 6\hat{k}$$\vec{BC} = \vec{c} - \vec{b} = (5-1)\hat{i} + (5-3)\hat{j} + (5-1)\hat{k} = 4\hat{i} + 2\hat{j} + 4\hat{k}$The area of $	riangle ABC$ is given by $\frac{1}{2} |\vec{AB} 	imes \vec{AC}|$.$\vec{AB} 	imes \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ -2 & -1 & 2 \ 2 & 1 & 6 \end{vmatrix} = \hat{i}(-6-2) - \hat{j}(-12-4) + \hat{k}(-2+2) = -8\hat{i} + 16\hat{j} + 0\hat{k}$Magnitude $|\vec{AB} 	imes \vec{AC}| = \sqrt{(-8)^2 + 16^2 + 0^2} = \sqrt{64 + 256} = \sqrt{320} = 8\sqrt{5}$.Area $= \frac{1}{2} 	imes 8\sqrt{5} = 4\sqrt{5}$.Also, Area $= \frac{1}{2} 	imes |\vec{BC}| 	imes p$, where $p$ is the altitude from $A$ to $BC$.$|\vec{BC}| = \sqrt{4^2 + 2^2 + 4^2} = \sqrt{16+4+16} = \sqrt{36} = 6$.So, $4\sqrt{5} = \frac{1}{2} 	imes 6 	imes p \implies 4\sqrt{5} = 3p \implies p = \frac{4\sqrt{5}}{3}$.

If the position vectors of the vertices of $\triangle ABC$ are $3\hat{i}+4\hat{j}-\hat{k}$, $\hat{i}+3\hat{j}+\hat{k}$, and $5(\hat{i}+\hat{j}+\hat{k})$ respectively, then the magnitude of the altitude from $A$ onto the side $BC$ is

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