Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

The distance of a point $(x_1,y_1,z_1)$ from the plane $Ax+By+Cz+D=0$ is given by $\frac{|A{x}_1+B{y}_1+C{z}_1+D|}{\sqrt{A^2+B^2+C^2}}$.

The distance of the point $(1,-1,\lambda )$ from the plane $3x-4y-12z+13=0$ is:

$d_1=\frac{|3(1)-4(-1)-12(\lambda )+13|}{\sqrt{3^2+(-4{)}^2+(-12{)}^2}}=\frac{|20-12\lambda |}{13}$

The distance of the point $(-3,0,1)$ from the plane is:

$d_2=\frac{|3(-3)-4(0)-12(1)+13|}{13}=\frac{|-8|}{13}=\frac{8}{13}$

Since the points are equidistant from the plane, $d_1=d_2$:

$\frac{|20-12\lambda |}{13}=\frac{8}{13}$

$|20-12\lambda |=8$

This gives two possible cases:

$20-12\lambda =8\Rightarrow 12\lambda =12\Rightarrow \lambda =1$

$20-12\lambda =-8\Rightarrow 12\lambda =28\Rightarrow \lambda =\frac{7}{3}$

The sum of all possible values of $\lambda$ is $1+\frac{7}{3}=\frac{10}{3}$.

Answer: $\frac{10}{3}$