Explanation

(d)
The given sphere is
$x^{2} + y^{2} + z^{2} + 4x - 2y - 6z - 155 = 0$
Its centre is $( - 2,\ 1,\ 3)$ and radius $= \sqrt{4 + 1 + 9 + 155} = \sqrt{169} = 13$
Therefore, distance of centre $( - 2,\ 1,\ 3)$ from the plane $12x + 4y + 3z = 327$
[image:12_image8.emf]
$= \frac{|12( - 2) + 4(1) + 3(3) - 327|}{\sqrt{144 + 16 + 9}} = 26$
Hence, the shortest distance is 13