Show that the points $(1, 1, 1)$ and $(- 3, 0, 1)$ are equidistant from the plane $¯ ¯ ¯ r \cdot (3 ˆ i + 4 ˆ j - 12 ˆ k) + 13 = 0$

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Solution

Plane : $3 x + 4 y - 12 z + 13 = 0$
Distance of $(1, 1, 1) = ∣ ∣ ∣ ∣ \frac{3 (1) + 4 (1) - 12 (1) + 13}{\sqrt{3^{2} + 4^{2} + 12^{2}}} ∣ ∣ ∣ ∣ = \frac{7 + 1}{\sqrt{25 + 144}} = \frac{8}{13}$
Distance of $(- 3, 0, 1) = ∣ ∣ ∣ \frac{- 9 - 12 + 13}{13} ∣ ∣ ∣ = ∣ ∣ ∣ \frac{1 - 9}{13} ∣ ∣ ∣ = \frac{8}{13}$

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