Question

If a plane meets the coordinate axes in $A,B$ and $C,$ such that the centroid of $\Delta ABC$ is $(1,2,4)$ then the perpendicular distance from origin to the plane is

• A. 12
• B. $\frac{12}{\sqrt{21}}$
• C. 7
• D. $\frac{7}{\sqrt{21}}$

Answer 1

The correct option is B $\frac{12}{\sqrt{21}}$

Let the required equation of the plane be
$\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\cdots \left(i\right)$
Then,it meets the coordinate axes in
$A(a,0,0),B(0,b,0)$ and $C(0,0,c)$
So, centroid of$\Delta ABC=(\frac{a+0+0}{3},\frac{0+b+0}{3},\frac{0+0+c}{3})$
$\Rightarrow G(\frac{a}{3},\frac{b}{3},\frac{c}{3})$
$\therefore \frac{a}{3}=1,\frac{b}{3}=2$ and $\frac{c}{3}=4$
$\Rightarrow a=3,b=6,c=12$
Hence,the required equation of the plane is
$\frac{x}{3}+\frac{y}{6}+\frac{z}{12}=1$
$\Rightarrow 4x+2y+z=12$
Now dividing the plane equation by $\sqrt{21}$, we have
$x\frac{4}{\sqrt{21}}+y\frac{2}{\sqrt{21}}+z\frac{1}{\sqrt{21}}=\frac{12}{\sqrt{21}}$
On comparing with $x\cos \alpha +y\cos \beta +z\cos \gamma =P$
we have $P=\frac{12}{\sqrt{21}}$