Question

A plane meets the coordinate axes in A, B, C such that the centroid of triangle ABC is the point $(p,q,r)$. If the equation of the plane is $\frac{x}{p}+\frac{y}{q}+\frac{z}{r}=k$, then $k=$

• A. $1$
• B. $2$
• C. $3$
• D. none of these

Answer 1

The correct option is C $3$

Given that:

Centriod of the $△ABC$ is $(p,q,r).$

Equation of the plane in intercept form is

$\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$ where, $a,b,c$ are x-intercept , y-intercept, z-intercept respectively.

Centriod of the $△ABC$ is

$(\frac{a+0+0}{3},\frac{0+b+0}{3},\frac{0+0+c}{3})$

$(\frac{a}{3},\frac{b}{3},\frac{c}{3})$

So, $a=3p,b=3q$ and $c=3r$

Now,

Equation of the plane is $\frac{x}{3p}+\frac{y}{3q}+\frac{z}{3r}=1$

or, $\frac{x}{p}+\frac{y}{q}+\frac{z}{r}=3$

Therefore, $k=3$

Hence, C is the correct option.