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(A) Let the intercepts of the plane on the coordinate axes be $a, b, c$. Thus, the points are $A(a, 0, 0)$, $B(0, b, 0)$, and $C(0, 0, c)$.The centroi

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$\frac{3}{2}$

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(A) Let the intercepts of the plane on the coordinate axes be $a, b, c$. Thus, 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 that the centroid is $(1, r, r^2)$, we have:$\frac{a}{3} = 1 \Rightarrow a = 3$$\frac{b}{3} = r \Rightarrow b = 3r$$\frac{c}{3} = r^2 \Rightarrow c = 3r^2$The equation of the plane in intercept form is $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$.Substituting the values of $a, b, c$, we get $\frac{x}{3} + \frac{y}{3r} + \frac{z}{3r^2} = 1$.Since the plane passes through $(5, 5, -12)$, we have:$\frac{5}{3} + \frac{5}{3r} - \frac{12}{3r^2} = 1$Multiplying by $3r^2$, we get $5r^2 + 5r - 12 = 3r^2$, which simplifies to $2r^2 + 5r - 12 = 0$.Factoring the quadratic equation: $(2r - 3)(r + 4) = 0$.Thus, $r = \frac{3}{2}$ or $r = -4$.

$A$ plane meets the coordinate axes at the points $A, B, C$ respectively in such a way that the centroid of $\triangle ABC$ is $(1, r, r^2)$ for some real $r$. If the plane passes through the point $(5, 5, -12)$, then $r=$

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