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Find the equation of a plane that contains the line of intersection of the planes x + 2y + 3z = 2 and x - y + z = 3, and is at a distance 2/√3 from the point (3, 1, -1).

1. 5x - 11y + z = 17
2. √2x + y = 3√2 - 1
3. x + y + z = √3
4. x - √2y = 1 - √2

Correct answer: 5x - 11y + z = 17

Solution

Take (x+2y+3z-2)+lambda(x-y+z-3)=0. Setting its distance from (3,1,-1) equal to 2/sqrt3 yields lambda=-7/2, giving 5x-11y+z=17. This plane contains the line of intersection and is at the required distance.

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