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

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(B) Let the equation of the plane be $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$.The coordinates of the points are $A(a, 0, 0)$,$B(0, b, 0)$,and $C(

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$\frac{x-1}{2}=\frac{y-1}{2}=\frac{z-2}{1}$

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(B) Let the equation of the plane be $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$.The coordinates of the points are $A(a, 0, 0)$,$B(0, b, 0)$,and $C(0, 0, c)$.The centroid of $\Delta ABC$ 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 $(1, 1, 2)$,we have $\frac{a}{3} = 1$,$\frac{b}{3} = 1$,and $\frac{c}{3} = 2$.Thus,$a = 3$,$b = 3$,and $c = 6$.The equation of the plane is $\frac{x}{3} + \frac{y}{3} + \frac{z}{6} = 1$,which simplifies to $2x + 2y + z = 6$.The normal vector to the plane is $\vec{n} = 2\hat{i} + 2\hat{j} + \hat{k}$,which serves as the direction ratio of the line perpendicular to the plane.The line passes through the centroid $(1, 1, 2)$ with direction ratios $(2, 2, 1)$.Therefore,the equation of the line is $\frac{x-1}{2} = \frac{y-1}{2} = \frac{z-2}{1}$.

$A$ plane $P$ meets the coordinate axes at $A, B$ and $C$ respectively. The centroid of $\Delta ABC$ is given to be $(1, 1, 2)$. Then the equation of the line through this centroid and perpendicular to the plane $P$ is

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