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(C) The direction vector of the line joining $(4, 1, -3)$ and $(2, 4, 3)$ is given by $\vec{n} = (2-4, 4-1, 3-(-3)) = (-2, 3, 6)$.Since the plane $P$

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$5$

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(C) The direction vector of the line joining $(4, 1, -3)$ and $(2, 4, 3)$ is given by $\vec{n} = (2-4, 4-1, 3-(-3)) = (-2, 3, 6)$.Since the plane $P$ is perpendicular to this line,the normal vector to the plane is $\vec{n} = (-2, 3, 6)$.The equation of the plane passing through $(1, -1, -5)$ with normal vector $\vec{n} = (-2, 3, 6)$ is:$-2(x - 1) + 3(y + 1) + 6(z + 5) = 0$$-2x + 2 + 3y + 3 + 6z + 30 = 0$$-2x + 3y + 6z + 35 = 0$ or $2x - 3y - 6z = 35$.The distance of the point $(3, -2, 2)$ from the plane $2x - 3y - 6z - 35 = 0$ is given by the formula $d = \frac{|ax_0 + by_0 + cz_0 + d|}{\sqrt{a^2 + b^2 + c^2}}$.$d = \frac{|2(3) - 3(-2) - 6(2) - 35|}{\sqrt{2^2 + (-3)^2 + (-6)^2}}$$d = \frac{|6 + 6 - 12 - 35|}{\sqrt{4 + 9 + 36}}$$d = \frac{|-35|}{\sqrt{49}} = \frac{35}{7} = 5$.

Let $P$ be the plane passing through the point $(1, -1, -5)$ and perpendicular to the line joining the points $(4, 1, -3)$ and $(2, 4, 3)$. Then the distance of $P$ from the point $(3, -2, 2)$ is

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