A) 2
B) \[-2\]
C) 1
D) \[-1\]
Correct Answer: B
Solution :
Centre of sphere \[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}+2ux+2\upsilon y+2\omega z+d=0\]is \[(-u,-\upsilon ,\omega )\]. Given equation of first sphere is \[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}+6x-8y-2z=13\] ...(i) whose centre is (-3, 4,1) and equation of second sphere is \[{{x}^{2}}+{{y}^{2}}\text{+ }{{z}^{2}}-10x+4y-2\text{ }z=8\] .. .(ii) whose centre is (5, -2,1). Mid-point of (-3, 4,1) and (5, -2,1) is (1,1,1). Since, the plane passes through (1,1,1). \[\therefore \] \[2a-3a+4a+6=0\] \[\Rightarrow \] \[3a=-\text{ }6\Rightarrow a=-2\]
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