The shortest distance from the plane 12x + 4y | vedclass

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(C) The equation of the sphere is $x^2 + y^2 + z^2 + 4x - 2y - 6z - 155 = 0$.Comparing this with the general equation $x^2 + y^2 + z^2 + 2gx + 2fy + 2

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$13$

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(C) The equation of the sphere is $x^2 + y^2 + z^2 + 4x - 2y - 6z - 155 = 0$.Comparing this with the general equation $x^2 + y^2 + z^2 + 2gx + 2fy + 2hz + c = 0$,we get the center $C = (-g, -f, -h) = (-2, 1, 3)$.The radius $r$ is given by $\sqrt{g^2 + f^2 + h^2 - c} = \sqrt{(-2)^2 + 1^2 + (-3)^2 - (-155)} = \sqrt{4 + 1 + 9 + 155} = \sqrt{169} = 13$.The perpendicular distance $d$ from the center $C(-2, 1, 3)$ to the plane $12x + 4y + 3z - 327 = 0$ is calculated as:$d = \frac{|12(-2) + 4(1) + 3(3) - 327|}{\sqrt{12^2 + 4^2 + 3^2}} = \frac{|-24 + 4 + 9 - 327|}{\sqrt{144 + 16 + 9}} = \frac{|-338|}{\sqrt{169}} = \frac{338}{13} = 26$.The shortest distance from the plane to the sphere is $d - r = 26 - 13 = 13$.

The shortest distance from the plane $12x + 4y + 3z = 327$ to the sphere $x^2 + y^2 + z^2 + 4x - 2y - 6z = 155$ is

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