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

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(D) 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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$1, \frac{7}{3}$

Vedclass · Answer

(D) The distance $d$ of a point $(x_1, y_1, z_1)$ from the plane $Ax + By + Cz + D = 0$ is given by $d = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}}$.For the point $(1, 1, \lambda)$, the distance $d_1$ from the plane $3x + 4y - 12z + 13 = 0$ is:$d_1 = \frac{|3(1) + 4(1) - 12(\lambda) + 13|}{\sqrt{3^2 + 4^2 + (-12)^2}} = \frac{|3 + 4 - 12\lambda + 13|}{\sqrt{9 + 16 + 144}} = \frac{|20 - 12\lambda|}{13}$.For the point $(-3, 0, 1)$, the distance $d_2$ from the plane is:$d_2 = \frac{|3(-3) + 4(0) - 12(1) + 13|}{\sqrt{3^2 + 4^2 + (-12)^2}} = \frac{|-9 + 0 - 12 + 13|}{13} = \frac{|-8|}{13} = \frac{8}{13}$.Since the points are equidistant, $d_1 = d_2$, so $\frac{|20 - 12\lambda|}{13} = \frac{8}{13}$.This implies $|20 - 12\lambda| = 8$, which gives two cases:Case $1$: $20 - 12\lambda = 8 \implies 12\lambda = 12 \implies \lambda = 1$.Case $2$: $20 - 12\lambda = -8 \implies 12\lambda = 28 \implies \lambda = \frac{28}{12} = \frac{7}{3}$.Thus, the values of $\lambda$ are $1$ and $\frac{7}{3}$.

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

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