The shortest distance between the plane $12 x + 64 y + 3 z = 327$ and the sphere $x^{2} + y^{2} + z^{2} + 4 x - 2 y - 6 z = 155$ is

11 \frac{3}{4}

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13

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39

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26

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Solution

The correct option is B $13$
The shortest distance required is the perpendicular distance from the centre of the sphere to the plane.
consider the above image

Center of the sphere = $(- 2, 1, 3)$ .

Radius of the sphere $C Q = \sqrt{(- 2)^{2} + 1^{2} + 3^{2} - (- 155)}$
= $\sqrt{4 + 1 + 9 + 155}$

= $\sqrt{1} 69$

$C Q = 13$

Perpendicular distance formula is
$\frac{| (a x_{1} + b y_{1} + c z_{1} + d)}{\sqrt{(a^{2} + b^{2} + c^{2}) |}}$

= $\frac{12 (- 2) + 4 (1) + 3 (3) - 327}{\sqrt{12^{2} + 4^{2} + 3^{2}}}$

length of the perpendicular $C P = \frac{338}{13} = 26$

$P Q = C P - C Q = 26 - 13 = 13$

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12 x + 4 y + 3 z = 327

x^{2} + y^{2} + z^{2} + 4 x - 2 y - 6 z = 155

x^{2} + y^{2} + z^{2} = 64

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x^{2} + y^{2} + z^{2} = 64

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