Question

A plane meets the co-ordinate axes in A, B, C such that the centroid of the triangle ABC is the point \((1,r,r^2)\), then the equation of the plane is,

• A. \(x+ry+r^2z = 3r^2\)
• B. \(r^2x+ry+z = 3r^2\)
• C. \(x+ry+r^2z = 3\)
• D. \(r^2x+ry+z = 3\)

Hint:

The centroid of the intercept triangle gives the three intercepts directly.

Answer 1

The Correct Option is B

Step 1: Test the intercepts:
Check each option for the intercepts $3, 3r, 3r^2$. In option (B), put $y = z = 0$: $x = 3$. Put $x = z = 0$: $y = 3r$. Put $x = y = 0$: $z = 3r^2$.

Step 2: Centroid:
Average of the three intercept points is $\left(1, r, r^2\right)$, as required.

Step 3: Other options:
Option (A) gives $x$-intercept $3r^2$, which would put the centroid at $x = r^2$, not 1.

Final Answer:
Option (B) is the plane. \[ \boxed{r^2x+ry+z=3r^2} \]