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in Three-dimensional geometry by (36.3k points)

If the distance between the plane x – 2y + z = d and the plane containing the lines (x – 1)/2 = (y – 2)/3 = (z – 3)/4 and (x – 2)/3 = (y – 3)/4 = (z – 4)/1 is √6, then |d| is...

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Best answer

Let the equation of the any plane be

ax + by + cz + d = 0 ...(i)

Since Eqs. (i) containing the lines, so

2a + 3b + 4c = 0

and 3a + 4b + 5c = 0.

Therefore, 

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