Let the vectors
$(2+a+b) \hat{i}+(a+2 b+c) \hat{j}-(b+c) \hat{k},(1+b) \hat{i}+2 b \hat{j}-b \hat{k}$
and $(2+\mathrm{b}) \hat{\mathrm{i}}+2 \mathrm{~b} \hat{\mathrm{j}}+(\mathrm{I}-\mathrm{b}) \hat{\mathrm{k}} \mathrm{a}, \mathrm{b}, \mathrm{c}, \in \mathbf{R}$
be co-planar. Then which of the following is true?
Correct Option: 1
If the vectors are co-planar,
$\left|\begin{array}{ccc}\mathrm{a}+\mathrm{b}+2 & \mathrm{a}+2 \mathrm{~b}+\mathrm{c} & -\mathrm{b}-\mathrm{c} \\ \mathrm{b}+1 & 2 \mathrm{~b} & -\mathrm{b} \\ \mathrm{b}+2 & 2 \mathrm{~b} & 1-\mathrm{b}\end{array}\right|=0$
Now $\mathrm{R}_{3} \rightarrow \mathrm{R}_{3}-\mathrm{R}_{2}, \mathrm{R}_{1} \rightarrow \mathrm{R}_{1}-\mathrm{R}_{2}$
So $\left|\begin{array}{ccc}\mathrm{a}+1 & \mathrm{a}+\mathrm{c} & -\mathrm{c} \\ \mathrm{b}+1 & 2 \mathrm{~b} & -\mathrm{b} \\ 1 & 0 & 1\end{array}\right|=0$
$=(a+1) 2 b-(a+c)(2 b+1)-c(-2 b)$
$=2 a b+2 b-2 a b-a-2 b c-c+2 b c$
$=2 b-a-c=0$
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