A) 1
B) 2
C) 3
D) 5
Correct Answer: D
Solution :
\[({{\vec{n}}_{1}}\times {{\vec{n}}_{2}})\,.{{\vec{n}}_{3}}\,=0\] (think!) Where \[{{\vec{n}}_{1}}=\hat{i}-\hat{j}+2\hat{k};\] \[{{\vec{n}}_{2}}=2\hat{i}+\hat{j}+\hat{k};\] and \[{{\vec{n}}_{3}}=m\hat{i}+2\hat{j}+3\hat{k};\] \[\left| \begin{matrix} 2 & 1 & 1 \\ 1 & -1 & 2 \\ m & 2 & 3 \\ \end{matrix} \right|=0\Rightarrow \,m=5\]You need to login to perform this action.
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