Concept:1). The general form of the equation of the sphere is x2 + y2 + z2 + 2ux + 2vy + 2wz + d = 0 where u, v, w and d are constant.2). Centre of the sphere is given by (-u, -v, -w)3). Radius of the sphere is given by r = \(\displaystyle \sqrt{{u^2+v^2+w^2-d}} \)4). The perpendicular distance of a point (x1, y1, z1) from plane ax + by + cz = d is given by \(\displaystyle p=\mid \frac{ax_1+by_1+cz_1-d}{\sqrt{{a^2+b^2+c^2}}}\mid \)Calculation:The given equation of the sphere is x2 + y2 + z2 - 2x - 3y - 4z = 0.Centre of the sphere is given by (-u, -v, -w) ? (1, \(\frac{3}{2}\), 2)The given equation of the plane is 6x + 4y + 3z - 12 = 0If p is the perpendicular distance from the center of the sphere to the plane,Then \(\displaystyle p=\mid \frac{ax_1+by_1+cz_1-d}{\sqrt{{a^2+b^2+c^2}}}\mid \)? \(\displaystyle p=\mid \frac{6\times1+4\times\frac{3}{2}+3\times2-12}{\sqrt{{6^2+4^2+3^2}}}\mid \)? \(\displaystyle p=\mid \frac{6}{\sqrt{{61}}}\mid \)? p = 0.768? 0.5 < p < 1