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(B) The vector equation of the line passing through points $A = i - 2j + k$ and $B = -2j + 3k$ is given by $r = A + \lambda(B - A)$.$r = (i - 2j + k)

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$\frac{1}{5}(6i - 10j + 3k)$

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(B) The vector equation of the line passing through points $A = i - 2j + k$ and $B = -2j + 3k$ is given by $r = A + \lambda(B - A)$.$r = (i - 2j + k) + \lambda(-i + 2k) = (1 - \lambda)i - 2j + (1 + 2\lambda)k$ ... $(i)$The plane passes through the origin $(0, 0, 0)$,$P = 4j$,and $Q = 2i + k$. The normal vector $n$ to the plane is given by $n = P 	imes Q = (4j) 	imes (2i + k) = 8(j 	imes i) + 4(j 	imes k) = -8k + 4i = 4i - 8k$.The equation of the plane is $r \cdot (4i - 8k) = 0$ ... $(ii)$Substituting the point from $(i)$ into $(ii)$:$((1 - \lambda)i - 2j + (1 + 2\lambda)k) \cdot (4i - 8k) = 0$$4(1 - \lambda) - 8(1 + 2\lambda) = 0$$4 - 4\lambda - 8 - 16\lambda = 0$$-4 - 20\lambda = 0 \Rightarrow \lambda = -\frac{1}{5}$Substituting $\lambda = -\frac{1}{5}$ into $(i)$:$r = (i - 2j + k) - \frac{1}{5}(-i + 2k) = i - 2j + k + \frac{1}{5}i - \frac{2}{5}k = \frac{6}{5}i - 2j + \frac{3}{5}k = \frac{1}{5}(6i - 10j + 3k)$.

The position vector of the point in which the line joining the points $i - 2j + k$ and $3k - 2j$ cuts the plane through the origin and the points $4j$ and $2i + k$,is

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