The values of a for which the two points (1, | vedclass

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(C) Let the plane be $f(x, y, z) = 3x + 4y - 12z + 13 = 0$. Two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ lie on opposite sides of the plane if $

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$a  \frac{1}{3}$

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(C) Let the plane be $f(x, y, z) = 3x + 4y - 12z + 13 = 0$. Two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ lie on opposite sides of the plane if $f(x_1, y_1, z_1) \cdot f(x_2, y_2, z_2) For point $P_1 = (1, a, 1)$,$f(1, a, 1) = 3(1) + 4(a) - 12(1) + 13 = 4a + 4$. For point $P_2 = (-3, 0, a)$,$f(-3, 0, a) = 3(-3) + 4(0) - 12(a) + 13 = 4 - 12a$. We require $(4a + 4)(4 - 12a) Dividing by $16$,we get $(a + 1)(1 - 3a) Multiplying by $-1$ reverses the inequality: $(a + 1)(3a - 1) > 0$. The roots are $a = -1$ and $a = \frac{1}{3}$. The inequality holds for $a  \frac{1}{3}$.

The values of $a$ for which the two points $(1, a, 1)$ and $(-3, 0, a)$ lie on the opposite sides of the plane $3x + 4y - 12z + 13 = 0$ satisfy:

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