If a plane meets the coordinate axes at A, B, | vedclass

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

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$4x + 2y + z = 12$

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(B) Let the intercepts of the plane on the coordinate axes be $a, b$,and $c$. Thus,the coordinates of the points are $A(a, 0, 0)$,$B(0, b, 0)$,and $C(0, 0, c)$.The centroid of the triangle $ABC$ is given by $\left( \frac{a+0+0}{3}, \frac{0+b+0}{3}, \frac{0+0+c}{3} \right) = \left( \frac{a}{3}, \frac{b}{3}, \frac{c}{3} \right)$.Given that the centroid is $(1, 2, 4)$,we have:$\frac{a}{3} = 1 \implies a = 3$$\frac{b}{3} = 2 \implies b = 6$$\frac{c}{3} = 4 \implies c = 12$The intercept form of the equation of a plane is $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$.Substituting the values of $a, b$,and $c$,we get $\frac{x}{3} + \frac{y}{6} + \frac{z}{12} = 1$.Multiplying the entire equation by $12$,we get $4x + 2y + z = 12$.

If a plane meets the coordinate axes at $A, B$,and $C$ such that the centroid of the triangle $ABC$ is $(1, 2, 4)$,then the equation of the plane is:

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