Let 6x - 3y + 2z - 6 = 0 be the given plane. | vedclass

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(C) The equation of the plane is $6x - 3y + 2z = 6$. Dividing by $6$,we get $\frac{x}{1} + \frac{y}{-2} + \frac{z}{3} = 1$. Thus,the intercepts are $a

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$3p$

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(C) The equation of the plane is $6x - 3y + 2z = 6$. Dividing by $6$,we get $\frac{x}{1} + \frac{y}{-2} + \frac{z}{3} = 1$. Thus,the intercepts are $a = 1, b = -2, c = 3$. The normal vector to the plane is $\vec{n} = 6\hat{i} - 3\hat{j} + 2\hat{k}$. The magnitude is $|\vec{n}| = \sqrt{6^2 + (-3)^2 + 2^2} = \sqrt{36 + 9 + 4} = \sqrt{49} = 7$. The direction cosines are $l = \frac{6}{7}, m = -\frac{3}{7}, n = \frac{2}{7}$. The perpendicular distance $p$ from the origin to the plane $Ax + By + Cz + D = 0$ is $p = \frac{|D|}{\sqrt{A^2 + B^2 + C^2}}$. Here $p = \frac{|-6|}{7} = \frac{6}{7}$. Now,calculate $|al + bm + cn| = |(1)(\frac{6}{7}) + (-2)(-\frac{3}{7}) + (3)(\frac{2}{7})| = |\frac{6}{7} + \frac{6}{7} + \frac{6}{7}| = |\frac{18}{7}|$. Since $p = \frac{6}{7}$,we have $|al + bm + cn| = 3 	imes \frac{6}{7} = 3p$.

Let $6x - 3y + 2z - 6 = 0$ be the given plane. If $a, b, c$ are the intercepts made by the plane on $X, Y, Z$-axes respectively; $l, m, n$ are the direction cosines of a normal drawn to the plane and $p$ is the perpendicular distance from the origin to the plane,then $|al + bm + cn|=$

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