A plane meets the coordinate axes at A, B, C | vedclass

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(B) The intercept form of the equation of a plane meeting the coordinate axes at $A(a, 0, 0)$,$B(0, b, 0)$,and $C(0, 0, c)$ is given by:$\frac{x}{a} +

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$4x+2y+z=12$

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(B) The intercept form of the equation of a plane meeting the coordinate axes at $A(a, 0, 0)$,$B(0, b, 0)$,and $C(0, 0, c)$ is given by:$\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$ $\ldots$ $(i)$The centroid of the triangle $ABC$ with vertices $A(a, 0, 0)$,$B(0, b, 0)$,and $C(0, 0, c)$ is given by the formula:$\left( \frac{a+0+0}{3}, \frac{0+b+0}{3}, \frac{0+0+c}{3} \right) = \left( \frac{a}{3}, \frac{b}{3}, \frac{c}{3} \right)$Given that the centroid is $(1, 2, 4)$,we equate the coordinates:$\frac{a}{3} = 1 \Rightarrow a = 3$$\frac{b}{3} = 2 \Rightarrow b = 6$$\frac{c}{3} = 4 \Rightarrow c = 12$Substituting these values into Eq. $(i)$:$\frac{x}{3} + \frac{y}{6} + \frac{z}{12} = 1$Multiplying the entire equation by $12$ to clear the denominators:$4x + 2y + z = 12$

$A$ plane meets the coordinate axes at $A, B, C$ such that the centroid of the triangle $ABC$ is $(1, 2, 4)$. Then,the equation of the plane is

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