Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

Let the coordinates of points A, B, and C on the coordinate axes be $(a,0,0)$, $(0,b,0)$, and $(0,0,c)$ respectively.

The equation of the plane passing through these points is $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$.

The centroid of the triangle ABC is given by $\left(\frac{a}{3},\frac{b}{3},\frac{c}{3}\right)$.

Comparing this with the given centroid $\left(2,-\frac{2}{3},\frac{1}{2}\right)$, we obtain:

$\frac{a}{3}=2\Rightarrow a=6$

$\frac{b}{3}=-\frac{2}{3}\Rightarrow b=-2$

$\frac{c}{3}=\frac{1}{2}\Rightarrow c=\frac{3}{2}$

Substituting the values of $a$, $b$, and $c$ in the equation of the plane:

$\frac{x}{6}+\frac{y}{-2}+\frac{z}{\frac{3}{2}}=1$

$\frac{x}{6}-\frac{y}{2}+\frac{2z}{3}=1$

Multiplying the entire equation by 6:

$x-3y+4z=6$

$x-3y+4z-6=0$

The perpendicular distance $d$ from the origin $(0,0,0)$ to this plane is:

$d=\frac{|1(0)-3(0)+4(0)-6|}{\sqrt{1^2+(-3{)}^2+4^2}}$

$d=\frac{|-6|}{\sqrt{1+9+16}}=\frac{6}{\sqrt{26}}$