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(B) Let the plane meet the coordinate axes at points $A(a, 0, 0)$,$B(0, b, 0)$,and $C(0, 0, c)$.The intercept form of the equation of the plane is $\f

Vedclass · Accepted Answer

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

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(B) Let the plane meet the coordinate axes at points $A(a, 0, 0)$,$B(0, b, 0)$,and $C(0, 0, c)$.The intercept form of the equation of the plane is $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1 \dots (i)$.The centroid of $	riangle ABC$ with vertices $(a, 0, 0)$,$(0, b, 0)$,and $(0, 0, c)$ 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 $(1, 2, 3)$,we equate the coordinates:$\frac{a}{3} = 1 \Rightarrow a = 3$$\frac{b}{3} = 2 \Rightarrow b = 6$$\frac{c}{3} = 3 \Rightarrow c = 9$Substituting the values of $a, b,$ and $c$ into equation $(i)$,we get $\frac{x}{3} + \frac{y}{6} + \frac{z}{9} = 1$.

If a plane meets the coordinate axes at $A, B$ and $C$ in such a way that the centroid of $\triangle ABC$ is at the point $(1, 2, 3)$,then the equation of the plane is

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