A) ? 2
B) 2
C) ? 1
D) 1
Correct Answer: A
Solution :
\[{{S}_{1}}\equiv {{x}^{2}}+{{y}^{2}}+{{z}^{2}}+6x-8y-2z=13,\]\[{{C}_{1}}\equiv (-3,\,4,\,1)\,\] \[{{S}_{2}}\equiv {{x}^{2}}+{{y}^{2}}+{{z}^{2}}-10x+4y-2z=8,\]\[{{C}_{2}}\equiv (5,\,-2,\,1)\] So mid point of \[{{C}_{1}}{{C}_{2}}\] (say P) \[\equiv P\left( \frac{5-3}{2},\frac{4-2}{2},\frac{1+1}{2} \right)=P(1,\,1,\,1)\] Now the plane \[2ax-3ay+4az+6=0\] passes through the point P, So, \[2a(1)-3a(1)+4a(1)+6=0=2a-3a+4a+6=0\] Þ \[3a+6=0\]Þ \[3a=-6\Rightarrow a=-2\].
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