If the points (1, -1, lambda) and (-3, 0, 1) | vedclass

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(B) The distance 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 +

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$\frac{10}{3}$

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(B) The distance 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 that the points $(1, -1, \lambda)$ and $(-3, 0, 1)$ are equidistant from the plane $3x - 4y - 12z + 13 = 0$,we have:$\frac{|3(1) - 4(-1) - 12(\lambda) + 13|}{\sqrt{3^2 + (-4)^2 + (-12)^2}} = \frac{|3(-3) - 4(0) - 12(1) + 13|}{\sqrt{3^2 + (-4)^2 + (-12)^2}}$$|3 + 4 - 12\lambda + 13| = |-9 - 0 - 12 + 13|$$|20 - 12\lambda| = |-8|$$|20 - 12\lambda| = 8$This implies $20 - 12\lambda = 8$ or $20 - 12\lambda = -8$.Case $1$: $20 - 12\lambda = 8 \Rightarrow 12\lambda = 12 \Rightarrow \lambda = 1$.Case $2$: $20 - 12\lambda = -8 \Rightarrow 12\lambda = 28 \Rightarrow \lambda = \frac{28}{12} = \frac{7}{3}$.The sum of all possible values of $\lambda$ is $1 + \frac{7}{3} = \frac{10}{3}$.

If the points $(1, -1, \lambda)$ and $(-3, 0, 1)$ are equidistant from the plane $3x - 4y - 12z + 13 = 0$,then the sum of all possible values of $\lambda$ is

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