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(A) The intercept form of the equation of a plane is given by $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$,where $a, b, c$ are the intercepts on the

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$8/3$

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(A) The intercept form of the equation of a plane is given by $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$,where $a, b, c$ are the intercepts on the $x, y, z$ axes respectively.Given intercepts are $a = 8, b = 4, c = 4$.Substituting these values,the equation of the plane is $\frac{x}{8} + \frac{y}{4} + \frac{z}{4} = 1$.Multiplying by $8$,we get $x + 2y + 2z = 8$,or $x + 2y + 2z - 8 = 0$.The length of the perpendicular 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,$A = 1, B = 2, C = 2, D = -8$.Thus,$d = \frac{|-8|}{\sqrt{1^2 + 2^2 + 2^2}} = \frac{8}{\sqrt{1 + 4 + 4}} = \frac{8}{\sqrt{9}} = \frac{8}{3}$.

If for a plane,the intercepts on the coordinate axes are $8, 4, 4$,then the length of the perpendicular from the origin onto the plane is:

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