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$For what value(s) of a will the two points (1, a, 1) and (- 3, 0, a) lie on opposite sides of the plane 3 x + 4 y - 12 z$ $+ 13 = 0 ?$

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$a < - 1 or a > 1 / 3$

$a = 0 only$

$0 < a < 1$

$- 1 < a < 1$

answer is A.

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Detailed Solution

$We must have (3 + 4 a - 12 + 13) (- 9 - 12 a + 13) < 0$

$\begin{array}{ll} or (a + 1) (12 a - 4) > 0 \\ or a < - 1 or a > 1 / 3 \end{array}$

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