A plane meets the X, Y, Z-axes in A, B, C res | vedclass

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(D) Let the coordinates of the points be $A = (a, 0, 0)$,$B = (0, b, 0)$,and $C = (0, 0, c)$.Since the centroid of $	riangle ABC$ is given as $(\frac

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$\frac{90}{19}$

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(D) Let the coordinates of the points be $A = (a, 0, 0)$,$B = (0, b, 0)$,and $C = (0, 0, c)$.Since the centroid of $	riangle ABC$ is given as $(\frac{a}{3}, \frac{b}{3}, \frac{c}{3}) = (2, -3, 5)$,we have $a = 6$,$b = -9$,and $c = 15$.The equation of the plane in intercept form is $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$,which becomes $\frac{x}{6} - \frac{y}{9} + \frac{z}{15} = 1$.This can be rewritten as $\frac{x}{6} - \frac{y}{9} + \frac{z}{15} - 1 = 0$.The perpendicular distance $d$ from the origin $(0, 0, 0)$ to the plane $Ax + By + Cz + D = 0$ is given by $d = \frac{|D|}{\sqrt{A^2 + B^2 + C^2}}$.Here,$A = \frac{1}{6}$,$B = -\frac{1}{9}$,$C = \frac{1}{15}$,and $D = -1$.$d = \frac{|-1|}{\sqrt{(\frac{1}{6})^2 + (-\frac{1}{9})^2 + (\frac{1}{15})^2}} = \frac{1}{\sqrt{\frac{1}{36} + \frac{1}{81} + \frac{1}{225}}}$.Calculating the denominator: $\frac{1}{36} + \frac{1}{81} + \frac{1}{225} = \frac{225 + 100 + 36}{8100} = \frac{361}{8100}$.Thus,$d = \frac{1}{\sqrt{\frac{361}{8100}}} = \frac{90}{19}$.

$A$ plane meets the $X, Y, Z$-axes in $A, B, C$ respectively. If the centroid of the $\triangle ABC$ is $(2, -3, 5)$,then the perpendicular distance from the origin to the given plane is:

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