If the points (1, 1, mu) and (-3, 0, 1) are e | vedclass

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(B) The distance $d$ of a point $(x_1, y_1, z_1)$ from the plane $ax + by + cz + d = 0$ is given by $d = \frac{|ax_1 + by_1 + cz_1 + d|}{\sqrt{a^2 + b

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$1, \frac{7}{3}$

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(B) The distance $d$ of a point $(x_1, y_1, z_1)$ from the plane $ax + by + cz + d = 0$ is given by $d = \frac{|ax_1 + by_1 + cz_1 + d|}{\sqrt{a^2 + b^2 + c^2}}$.Given plane: $3x + 4y - 12z + 13 = 0$.Distance of $(1, 1, \mu)$ from the plane: $d_1 = \frac{|3(1) + 4(1) - 12(\mu) + 13|}{\sqrt{3^2 + 4^2 + (-12)^2}} = \frac{|20 - 12\mu|}{13}$.Distance of $(-3, 0, 1)$ from the plane: $d_2 = \frac{|3(-3) + 4(0) - 12(1) + 13|}{\sqrt{3^2 + 4^2 + (-12)^2}} = \frac{|-9 - 12 + 13|}{13} = \frac{|-8|}{13} = \frac{8}{13}$.Since the points are equidistant, $d_1 = d_2$, so $\frac{|20 - 12\mu|}{13} = \frac{8}{13}$.$|20 - 12\mu| = 8$.Case $1$: $20 - 12\mu = 8 \implies 12\mu = 12 \implies \mu = 1$.Case $2$: $20 - 12\mu = -8 \implies 12\mu = 28 \implies \mu = \frac{28}{12} = \frac{7}{3}$.Thus, the values of $\mu$ are $1, \frac{7}{3}$.

If the points $(1, 1, \mu)$ and $(-3, 0, 1)$ are equidistant from the plane $\vec{r} \cdot (3\hat{i} + 4\hat{j} - 12\hat{k}) + 13 = 0$, then the values of $\mu$ are

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