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

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(C) 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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$3\lambda^2 - 10\lambda + 7 = 0$

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(C) The distance of a point $(x_1, y_1, z_1)$ from the plane $Ax + By + Cz + D = 0$ is given by $d = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}}$.For point $(1, 1, \lambda)$,the distance is $d_1 = \frac{|3(1) + 4(1) - 12(\lambda) + 13|}{\sqrt{3^2 + 4^2 + (-12)^2}} = \frac{|20 - 12\lambda|}{\sqrt{9 + 16 + 144}} = \frac{|20 - 12\lambda|}{13}$.For point $(-3, 0, 1)$,the distance is $d_2 = \frac{|3(-3) + 4(0) - 12(1) + 13|}{\sqrt{3^2 + 4^2 + (-12)^2}} = \frac{|-9 - 12 + 13|}{13} = \frac{|-8|}{13} = \frac{8}{13}$.Since the points are equidistant,$d_1 = d_2$,so $\frac{|20 - 12\lambda|}{13} = \frac{8}{13}$.This implies $|20 - 12\lambda| = 8$,or $|5 - 3\lambda| = 2$.Squaring both sides,$(5 - 3\lambda)^2 = 2^2$,which gives $25 - 30\lambda + 9\lambda^2 = 4$.Rearranging the terms,$9\lambda^2 - 30\lambda + 21 = 0$.Dividing by $3$,we get $3\lambda^2 - 10\lambda + 7 = 0$.

If the points $(1, 1, \lambda )$ and $(-3, 0, 1)$ are equidistant from the plane $3x + 4y - 12z + 13 = 0,$ then $\lambda$ satisfies the equation

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