Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

The equation of the given plane in Cartesian form is $3x+4y-12z+13=0$.

The perpendicular 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 $(1,1,\mu )$ from the plane is:

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

The distance of the point $(-3,0,1)$ from the plane 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 from the plane, $d_1=d_2$.

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

$|20-12\mu |=8$

This gives two cases:

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

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

The values of $\mu$ are $1,\frac{7}{3}$.

Answer: $1,\frac{7}{3}$