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Find the distance of the point (1, -2, 3) from the plane x - y + z = 5, measured parallel to the line x/2 = y/3 = (z-3)/(-4).

1. (1/5) * sqrt(21)
2. (1/5) * sqrt(29)
3. (1/5) * sqrt(13)
4. 2 / sqrt(5)

Correct answer: (1/5) * sqrt(29)

Solution

Direction ratios of the given line: (2, 3, -4). Line through P(1,-2,3) parallel to this direction: x = 1+2t, y = -2+3t, z = 3-4t. Substitute into the plane x - y + z = 5: (1+2t) - (-2+3t) + (3-4t) = 5. 1+2t+2-3t+3-4t = 5. 6 - 5t = 5. -5t = -1. t = 1/5. Point of intersection: x = 1+2/5 = 7/5, y = -2+3/5 = -7/5, z = 3-4/5 = 11/5. Distance from P(1,-2,3) to this point: sqrt((7/5-1)² + (-7/5+2)² + (11/5-3)²) = sqrt((2/5)² + (3/5)² + (-4/5)²) = sqrt(4+9+16)/5 = sqrt(29)/5.

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