A plane meets the coordinate axes at A, B, C | vedclass

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(D) Let the equation of the plane be $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$, where $a, b, c$ are the $x, y, z$ intercepts respectively. The coo

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$15x + 10y + 6z = 90$

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(D) Let the equation of the plane be $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$, where $a, b, c$ are the $x, y, z$ intercepts respectively. The coordinates of the vertices are $A(a, 0, 0)$, $B(0, b, 0)$, and $C(0, 0, c)$. The centroid of $	riangle ABC$ is given by $\left(\frac{a+0+0}{3}, \frac{0+b+0}{3}, \frac{0+0+c}{3}ight) = \left(\frac{a}{3}, \frac{b}{3}, \frac{c}{3}ight)$. Given that the centroid is $(2, 3, 5)$, we have: $\frac{a}{3} = 2 \implies a = 6$ $\frac{b}{3} = 3 \implies b = 9$ $\frac{c}{3} = 5 \implies c = 15$ Substituting these values into the intercept form of the plane equation: $\frac{x}{6} + \frac{y}{9} + \frac{z}{15} = 1$ To simplify, multiply the entire equation by the least common multiple of $6, 9, 15$, which is $90$: $15x + 10y + 6z = 90$.

$A$ plane meets the coordinate axes at $A, B, C$ respectively such that the centroid of the $\triangle ABC$ is $(2, 3, 5)$. Then, the equation of that plane is

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