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

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$x+y+2z=18$

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(C) Let the equation of the plane be $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$. Since the plane cuts the coordinate 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: $\frac{a}{3} = 6 \implies a = 18$ $\frac{b}{3} = 6 \implies b = 18$ $\frac{c}{3} = 3 \implies c = 9$ Substituting these values into the intercept form of the plane equation: $\frac{x}{18} + \frac{y}{18} + \frac{z}{9} = 1$ Multiplying the entire equation by $18$, we get: $x + y + 2z = 18$.

If a plane cuts the coordinate axes at $A, B$ and $C$ respectively such that the centroid of the triangle $ABC$ is $(6, 6, 3)$, then find the equation of that plane.

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