A plane meets the coordinate axes at points A | vedclass

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(B) Let the intercepts of the plane on the $x, y, z$ axes be $a, b, c$ respectively. The coordinates of the points are $A(a, 0, 0)$, $B(0, b, 0)$, and

Vedclass · Accepted Answer

$r^2x + ry + z = 3r^2$

Vedclass · Answer

(B) Let the intercepts of the plane on the $x, y, z$ axes be $a, b, c$ respectively. The coordinates of the points are $A(a, 0, 0)$, $B(0, b, 0)$, and $C(0, 0, c)$.The centroid of $	riangle ABC$ is given by $(\frac{a}{3}, \frac{b}{3}, \frac{c}{3})$.Given the centroid is $(1, r, r^2)$, we have $\frac{a}{3} = 1 \implies a = 3$, $\frac{b}{3} = r \implies b = 3r$, and $\frac{c}{3} = r^2 \implies c = 3r^2$.The intercept form of the equation of a plane is $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$.Substituting the values: $\frac{x}{3} + \frac{y}{3r} + \frac{z}{3r^2} = 1$.Multiplying the entire equation by $3r^2$, we get $r^2x + ry + z = 3r^2$.

$A$ plane meets the coordinate axes at points $A, B, C$ such that the centroid of the triangle $ABC$ is $(1, r, r^2)$. Find the equation of the plane.

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