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

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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 the points $A(1, 1, \lambda)$ and $B(-3, 0, 1)$ are equidistant from the plane $3x + 4y - 12z + 13 = 0$.Distance from $A$: $d_1 = \frac{|3(1) + 4(1) - 12(\lambda) + 13|}{\sqrt{3^2 + 4^2 + (-12)^2}} = \frac{|20 - 12\lambda|}{\sqrt{9 + 16 + 144}} = \frac{|20 - 12\lambda|}{13}$.Distance from $B$: $d_2 = \frac{|3(-3) + 4(0) - 12(1) + 13|}{\sqrt{3^2 + 4^2 + (-12)^2}} = \frac{|-9 + 0 - 12 + 13|}{13} = \frac{|-8|}{13} = \frac{8}{13}$.Since $d_1 = d_2$,we have $\frac{|20 - 12\lambda|}{13} = \frac{8}{13}$,which implies $|20 - 12\lambda| = 8$.Case $1$: $20 - 12\lambda = 8 \Rightarrow 12\lambda = 12 \Rightarrow \lambda = 1$.Case $2$: $20 - 12\lambda = -8 \Rightarrow 12\lambda = 28 \Rightarrow \lambda = \frac{28}{12} = \frac{7}{3}$.Since $\lambda$ must be an integer,the value is $\lambda = 1$.

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

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