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(C) The equation of the line passing through $P(1, -2, 3)$ and parallel to the line $\frac{x}{1} = \frac{y}{4} = \frac{z}{5}$ is given by $\frac{x-1}{

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$2\sqrt{42}$

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(C) The equation of the line passing through $P(1, -2, 3)$ and parallel to the line $\frac{x}{1} = \frac{y}{4} = \frac{z}{5}$ is given by $\frac{x-1}{1} = \frac{y+2}{4} = \frac{z-3}{5} = \lambda$.Any point $F$ on this line can be represented as $(\lambda+1, 4\lambda-2, 5\lambda+3)$.Since $F$ lies on the plane $2x + 3y - 4z + 22 = 0$,we substitute the coordinates of $F$ into the plane equation:$2(\lambda+1) + 3(4\lambda-2) - 4(5\lambda+3) + 22 = 0$$2\lambda + 2 + 12\lambda - 6 - 20\lambda - 12 + 22 = 0$$-6\lambda + 6 = 0 \Rightarrow \lambda = 1$.Substituting $\lambda = 1$ back into the coordinates of $F$,we get $F(2, 2, 8)$.Since $F$ is the midpoint of $PQ$,the distance $PQ = 2PF$.The distance $PF = \sqrt{(2-1)^2 + (2-(-2))^2 + (8-3)^2} = \sqrt{1^2 + 4^2 + 5^2} = \sqrt{1 + 16 + 25} = \sqrt{42}$.Therefore,$PQ = 2PF = 2\sqrt{42}$.

If the image of the point $P(1, -2, 3)$ in the plane $2x + 3y - 4z + 22 = 0$ measured parallel to the line $\frac{x}{1} = \frac{y}{4} = \frac{z}{5}$ is $Q$,then $PQ$ is equal to:

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