If (bar{i}+bar{j}+bar{k}),(bar{i}+2bar{j}+3ba | vedclass

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(C) Let the position vectors of vertices $A, B, C$ be $\vec{a} = \bar{i}+\bar{j}+\bar{k}$,$\vec{b} = \bar{i}+2\bar{j}+3\bar{k}$,and $\vec{c} = 2\bar{i

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$\bar{r}=\bar{i}+\bar{j}+\bar{k}+t(\bar{i}-\bar{j}+2\bar{k})$

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(C) Let the position vectors of vertices $A, B, C$ be $\vec{a} = \bar{i}+\bar{j}+\bar{k}$,$\vec{b} = \bar{i}+2\bar{j}+3\bar{k}$,and $\vec{c} = 2\bar{i}-\bar{j}+\bar{k}$.The altitude through $A$ is perpendicular to the side $BC$.The vector along $BC$ is $\vec{BC} = \vec{c} - \vec{b} = (2-1)\bar{i} + (-1-2)\bar{j} + (1-3)\bar{k} = \bar{i} - 3\bar{j} - 2\bar{k}$.Let $\vec{n}$ be the direction vector of the altitude through $A$. Since the altitude is perpendicular to $BC$,$\vec{n}$ must be perpendicular to $\vec{BC}$.Also,the altitude lies in the plane of $	riangle ABC$,so it is perpendicular to the normal of the plane,$\vec{N} = \vec{AB} 	imes \vec{AC}$.$\vec{AB} = \vec{b} - \vec{a} = 0\bar{i} + \bar{j} + 2\bar{k}$.$\vec{AC} = \vec{c} - \vec{a} = \bar{i} - 2\bar{j} + 0\bar{k}$.$\vec{N} = \vec{AB} 	imes \vec{AC} = \begin{vmatrix} \bar{i} & \bar{j} & \bar{k} \ 0 & 1 & 2 \ 1 & -2 & 0 \end{vmatrix} = \bar{i}(0 - (-4)) - \bar{j}(0 - 2) + \bar{k}(0 - 1) = 4\bar{i} + 2\bar{j} - \bar{k}$.The direction of the altitude is $\vec{v} = \vec{N} 	imes \vec{BC} = \begin{vmatrix} \bar{i} & \bar{j} & \bar{k} \ 4 & 2 & -1 \ 1 & -3 & -2 \end{vmatrix} = \bar{i}(-4 - 3) - \bar{j}(-8 - (-1)) + \bar{k}(-12 - 2) = -7\bar{i} + 7\bar{j} - 14\bar{k}$.Simplifying the direction vector by dividing by $-7$,we get $\bar{i} - \bar{j} + 2\bar{k}$.The equation of the line passing through $A(\bar{a})$ with direction $\vec{v}$ is $\bar{r} = \vec{a} + t\vec{v} = (\bar{i}+\bar{j}+\bar{k}) + t(\bar{i}-\bar{j}+2\bar{k})$.Thus,the correct option is $C$.

If $(\bar{i}+\bar{j}+\bar{k})$,$(\bar{i}+2\bar{j}+3\bar{k})$ and $(2\bar{i}-\bar{j}+\bar{k})$ are the position vectors of the vertices $A$,$B$ and $C$ of $\triangle ABC$ respectively,then the vector equation of the altitude through $A$ is

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