JEE Main · MathematicsMedium

$The distance of the point (1, - 2, 3) from the plane passing through the points (1, 2, 3), (- 1, 3, 0), and (2, 1, - 1) is:$

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Correct answer: A — $\frac{15}{107}$

Solution

The normal vector is n = ((- 1 - 1), (3 - 2), (0 - 3)) \times ((2 - 1), (1 - 2), (- 1 - 3)) = (- 2, 1, - 3) \times (1, - 1, - 4) = (- 7, - 11, 1) . The plane equation is - 7 (x - 1) - 11 (y - 2) + 1 (z - 3) = 0, or $7x + 11y - z - 26 = 0$. Distance is ∣7 (1) + 11 (- 2) - 3 - 26∣/ 49 + 121 + 1 = ∣ - 44∣/ 171 .

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