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

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$\frac{12}{\sqrt{29}}$

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(C) The intercept form of the equation of a plane is $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$,where $a, b, c$ are the intercepts on the $x, y, z$ axes respectively.Given intercepts are $a = -6, b = 3, c = 4$.Substituting these values,we get $\frac{x}{-6} + \frac{y}{3} + \frac{z}{4} = 1$.To simplify,multiply by the least common multiple of $6, 3, 4$,which is $12$:$-2x + 4y + 3z = 12$,or $-2x + 4y + 3z - 12 = 0$.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,$A = -2, B = 4, C = 3, D = -12$.$d = \frac{|-12|}{\sqrt{(-2)^2 + 4^2 + 3^2}} = \frac{12}{\sqrt{4 + 16 + 9}} = \frac{12}{\sqrt{29}}$.

If a plane cuts off intercepts $-6, 3, 4$ from the coordinate axes,then the length of the perpendicular from the origin to the plane is

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