A) \[{{a}^{2}}-9a+18=0\]
B) \[{{a}^{2}}-6a-12=0\]
C) \[{{a}^{2}}-6a-18=0\]
D) \[{{a}^{2}}-9a+12=0\]
Correct Answer: C
Solution :
Case I Let line \[{{l}_{1}}\equiv x-3y=p\] and \[{{l}_{2}}\equiv ax+2y=p\]are perpendicular, then \[\frac{1}{3}\times -\frac{a}{2}=-1\Rightarrow a=6\] Case II Let line \[{{l}_{2}}\equiv ax+2y=p\] and \[{{l}_{3}}\equiv ax+y=r\] are perpendicular, then \[\frac{-a}{2}\times -a=-1\Rightarrow {{a}^{2}}=-2\] (not possible) Case III Let line \[{{l}_{3}}\equiv ax+y=r\] and \[{{l}_{1}}\equiv x-3y=p\] are perpendicular, then \[-a\times \frac{1}{3}=-1\Rightarrow a=3\] \[\therefore \] Formation of quadratic equation in a. whose roots are 3 and 6. \[{{a}^{2}}-(6+3)a+(6\cdot 3)=0\Rightarrow {{a}^{2}}-9a+18=0\]
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