A plane cuts the coordinate axes X, Y, Z at A | vedclass

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(D) Let the equation of the plane be $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$. Since the plane cuts the axes at $A(a, 0, 0)$,$B(0, b, 0)$,and $C(

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$x+y+2z-18=0$

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(D) Let the equation of the plane be $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$. Since the plane cuts the axes at $A(a, 0, 0)$,$B(0, b, 0)$,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 $(6, 6, 3)$,we have: $\frac{a}{3} = 6 \Rightarrow a = 18$ $\frac{b}{3} = 6 \Rightarrow b = 18$ $\frac{c}{3} = 3 \Rightarrow c = 9$ Substituting these values into the plane equation: $\frac{x}{18} + \frac{y}{18} + \frac{z}{9} = 1$ Multiplying by $18$,we get: $x + y + 2z = 18$ $x + y + 2z - 18 = 0$.

$A$ plane cuts the coordinate axes $X, Y, Z$ at $A, B, C$ respectively such that the centroid of the $\triangle ABC$ is $(6, 6, 3)$. Then the equation of that plane is

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