If a plane meets the axes X, Y, Z in A, B, C | vedclass

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(C) Let the intercepts of the plane on the $X, Y, Z$ axes be $a, b, c$ respectively. Thus,the coordinates of the points are $A=(a, 0, 0)$,$B=(0, b, 0)

Vedclass · Accepted Answer

$\frac{x}{3}+\frac{y}{6}+\frac{z}{9}=1$

Vedclass · Answer

(C) Let the intercepts of the plane on the $X, Y, Z$ axes be $a, b, c$ respectively. Thus,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 $(\frac{a+0+0}{3}, \frac{0+b+0}{3}, \frac{0+0+c}{3}) = (1, 2, 3)$. Equating the coordinates,we get: $\frac{a}{3} = 1 \Rightarrow a = 3$ $\frac{b}{3} = 2 \Rightarrow b = 6$ $\frac{c}{3} = 3 \Rightarrow c = 9$ The intercept form of the equation of a plane is $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$. Substituting the values of $a, b, c$,we get the equation of the plane as $\frac{x}{3} + \frac{y}{6} + \frac{z}{9} = 1$.

If a plane meets the axes $X, Y, Z$ in $A, B, C$ respectively such that the centroid of $\triangle ABC$ is $(1, 2, 3)$,then the equation of the plane is

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