If $(1, 1, K)$ and $(- 3, 0, 1)$ are at equal perpendicular distance form $3 x + 4 y - 12 x = - 12$ find k.

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Solution

$(1, 1, K); (- 3, 0, 1)$
$3 x + 4 y - 12 z + 12 = 0$
from question
$\frac{3 (1) + 4 (1) - 12 (k) + 12}{\sqrt{9 + 16 + 144}} = \frac{3 (- 3) + 4 (0) - 12 (1) + 12}{\sqrt{9 + 16 + 144}}$
$3 + 4 - 12 k + 12 = - 9 - 12 + 12$
$- 12 k + 19 = - 9$
$k = \frac{7}{3}$

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