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(C) Let $Q$ be the image of the point $P(\vec{i} + 3\vec{k})$ in the plane $\vec{r} \cdot (\vec{i} + \vec{j} + \vec{k}) = 1$. Then the line $PQ$ is no

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$-\vec{i} - 2\vec{j} + \vec{k}$

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(C) Let $Q$ be the image of the point $P(\vec{i} + 3\vec{k})$ in the plane $\vec{r} \cdot (\vec{i} + \vec{j} + \vec{k}) = 1$. Then the line $PQ$ is normal to the plane.Since $PQ$ passes through $P$ and is normal to the given plane,the equation of line $PQ$ is $\vec{r} = (\vec{i} + 3\vec{k}) + \lambda(\vec{i} + \vec{j} + \vec{k})$.Since $Q$ lies on the line $PQ$,let the position vector of $Q$ be $(1 + \lambda)\vec{i} + \lambda\vec{j} + (3 + \lambda)\vec{k}$.Let $R$ be the midpoint of $PQ$. The position vector of $R$ is $\frac{(\vec{i} + 3\vec{k}) + ((1 + \lambda)\vec{i} + \lambda\vec{j} + (3 + \lambda)\vec{k})}{2} = (\frac{\lambda + 2}{2})\vec{i} + (\frac{\lambda}{2})\vec{j} + (\frac{\lambda + 6}{2})\vec{k}$.Since $R$ lies on the plane $\vec{r} \cdot (\vec{i} + \vec{j} + \vec{k}) = 1$,we have:$(\frac{\lambda + 2}{2} + \frac{\lambda}{2} + \frac{\lambda + 6}{2}) = 1$$\frac{3\lambda + 8}{2} = 1$$3\lambda + 8 = 2$$3\lambda = -6 \implies \lambda = -2$.Substituting $\lambda = -2$ into the expression for $Q$:$Q = (1 - 2)\vec{i} + (-2)\vec{j} + (3 - 2)\vec{k} = -\vec{i} - 2\vec{j} + \vec{k}$.

The image of the point with position vector $\vec{i} + 3\vec{k}$ in the plane $\vec{r} \cdot (\vec{i} + \vec{j} + \vec{k}) = 1$ is

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