Let the line passing through the points P(2, | vedclass

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

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$3$

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(D) The equation of the line passing through $P(2, -1, 2)$ and $Q(5, 3, 4)$ is $\frac{x - 2}{5 - 2} = \frac{y - (-1)}{3 - (-1)} = \frac{z - 2}{4 - 2}$,which simplifies to $\frac{x - 2}{3} = \frac{y + 1}{4} = \frac{z - 2}{2} = \lambda$. Any point on this line is $R(3\lambda + 2, 4\lambda - 1, 2\lambda + 2)$. Since $R$ lies on the plane $x - y + z = 4$,we have $(3\lambda + 2) - (4\lambda - 1) + (2\lambda + 2) = 4$. Solving for $\lambda$: $\lambda + 5 = 4 \implies \lambda = -1$. Thus,$R = (3(-1) + 2, 4(-1) - 1, 2(-1) + 2) = (-1, -5, 0)$. We need the distance of $R(-1, -5, 0)$ from the plane $x + 2y + 3z + 2 = 0$ measured parallel to the line $\frac{x - 7}{2} = \frac{y + 3}{2} = \frac{z - 2}{1}$. The line passing through $R$ parallel to the given line is $\frac{x + 1}{2} = \frac{y + 5}{2} = \frac{z - 0}{1} = k$. Any point on this line is $T(2k - 1, 2k - 5, k)$. Since $T$ lies on the plane $x + 2y + 3z + 2 = 0$,we have $(2k - 1) + 2(2k - 5) + 3(k) + 2 = 0$. $2k - 1 + 4k - 10 + 3k + 2 = 0 \implies 9k - 9 = 0 \implies k = 1$. Thus,$T = (2(1) - 1, 2(1) - 5, 1) = (1, -3, 1)$. The distance $RT = \sqrt{(1 - (-1))^2 + (-3 - (-5))^2 + (1 - 0)^2} = \sqrt{2^2 + 2^2 + 1^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3$.

Let the line passing through the points $P(2, -1, 2)$ and $Q(5, 3, 4)$ meet the plane $x - y + z = 4$ at the point $R$. Then the distance of the point $R$ from the plane $x + 2y + 3z + 2 = 0$ measured parallel to the line $\frac{x - 7}{2} = \frac{y + 3}{2} = \frac{z - 2}{1}$ is equal to

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