A) 2
B) 0
C) -1
D) 1
Correct Answer: C
Solution :
Let \[\vec{a}=\mu \hat{i}+\hat{j}+\hat{k},\]\[\hat{b}=\hat{i}+\mu \hat{j}+\hat{k}\]and\[\vec{c}=\hat{i}+\hat{j}+\mu \hat{k}\] \[\therefore \]\[\vec{a}\,\vec{b}\,\vec{c}=\left| \begin{matrix} \mu & 1 & 1 \\ 1 & \mu & 1 \\ 1 & 1 & \mu \\ \end{matrix} \right|\] \[=\mu ({{\mu }^{2}}-1)-1(\mu -1)+1(1-\mu )\] \[={{\mu }^{3}}-\mu -\mu +1+1-\mu ={{\mu }^{3}}-3\mu +2\] \[\vec{a},\vec{b}\]and \[\vec{c}\]are coplanar, so \[[\vec{a}\,\vec{b}\,\vec{c}]=0\] \[\therefore \]\[{{\mu }^{3}}-3\mu +2=0\Rightarrow {{(\mu -1)}^{2}}(\mu +2)=0\] \[\Rightarrow \mu =1,\,1,-2\] \[\therefore \]The sum of the distinct real values of \[\mu =1-2=-1.\]
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