The position vectors of two points P and Q ar | vedclass

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(C) Given position vectors of points $P$ and $Q$ are $\vec{p} = 3i + j + 2k$ and $\vec{q} = i - 2j - 4k$. The vector $\overrightarrow{PQ} = \vec{q} -

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$r \cdot (2i + 3j + 6k) = -28$

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(C) Given position vectors of points $P$ and $Q$ are $\vec{p} = 3i + j + 2k$ and $\vec{q} = i - 2j - 4k$. The vector $\overrightarrow{PQ} = \vec{q} - \vec{p} = (i - 2j - 4k) - (3i + j + 2k) = -2i - 3j - 6k$. The plane passes through $Q$ and is perpendicular to $\overrightarrow{PQ}$. The normal vector to the plane is $\vec{n} = \overrightarrow{PQ} = -2i - 3j - 6k$. The equation of the plane is given by $(r - \vec{q}) \cdot \vec{n} = 0$,which is $r \cdot \vec{n} = \vec{q} \cdot \vec{n}$. $\vec{q} \cdot \vec{n} = (i - 2j - 4k) \cdot (-2i - 3j - 6k) = (1)(-2) + (-2)(-3) + (-4)(-6) = -2 + 6 + 24 = 28$. Thus,$r \cdot (-2i - 3j - 6k) = 28$,which can be written as $r \cdot (2i + 3j + 6k) = -28$.

The position vectors of two points $P$ and $Q$ are $3i + j + 2k$ and $i - 2j - 4k$ respectively. The equation of the plane passing through $Q$ and perpendicular to $PQ$ is:

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