A) For no value of \[\lambda \operatorname{and}\theta \]
B) For all values of \[\lambda \]and\[\theta \]
C) For all values of\[\lambda \], and only two values of\[\theta \]
D) For only one value of \[\lambda \] and all values of \[\theta \]
Correct Answer: A
Solution :
determinant of coefficients \[=\left| \begin{matrix} \lambda & -1 & \cos \theta \\ 3 & 1 & 2 \\ \cos \theta & 1 & 2 \\ \end{matrix} \right|=\cos \theta -{{\cos }^{2}}\theta +6\] and this is positive for all \[\theta \] since \[\left| \cos \theta \right|\le 1.\]the only solution is therefore the trivial solution.
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