Let $\hat{i},\hat{j}$ and $\hat{k}$ be the unit vectors along the three positive coordinate axes. Let

$\overrightarrow{a}=3\hat{i}+\hat{j}-\hat{k}$

$\overrightarrow{b}=\hat{i}+b_2\hat{j}+b_3\hat{k}, b_2,b_3\in \mathrm{ℝ}$

$\overrightarrow{c}=c_1\hat{i}+c_2\hat{j}+c_3\hat{k},c_1,c_2,c_3\in \mathrm{ℝ}$

be three vectors such that $b_2{b}_3>0,\overrightarrow{a}·\overrightarrow{b}=0$ and $\left(\begin{array}{ccc}0& -c_3& c_2\\ c_3& 0& -c_1\\ -c_2& c_1& 0\end{array}\right)\left(\begin{array}{l}1\\ b_2\\ b_3\end{array}\right)=\left(\begin{array}{c}3-c_1\\ 1-c_2\\ -1-c_3\end{array}\right).$ Then, which of the following is/are TRUE?

Given,

$\overrightarrow{a}=3\hat{i}+\hat{j}-\hat{k}$

$\overrightarrow{b}=\hat{i}+b_2\hat{j}+b_3\hat{k}$

$\overrightarrow{c}=c_1\hat{i}+c_2\hat{j}+c_3\hat{k}$

$\left[\begin{array}{ccc}0& -c_3& c_2\\ c_3& 0& -c_1\\ -c_2& c_1& 0\end{array}\right]\left[\begin{array}{c}1\\ b_2\\ b_3\end{array}\right]=\left[\begin{array}{c}3-c_1\\ 1-c_2\\ -1-c_3\end{array}\right]$

Now multiply and compare we get,

$b_2{c}_3-b_3{c}_2=c_1-3 .......\left(1\right)$

$c_3-b_3{c}_1=1-c_2...........\left(2\right)$

$c_2-b_2{c}_1=1+c_3...........\left(3\right)$

Now applying operation $\text{equation }\left(1\right)\hat{i}-\left(2\right)\hat{j}+\left(3\right)\hat{k}$ we get,

$\hat{i}\left(b_2{c}_3-c_2{b}_3\right)-\hat{j}\left(c_3-b_3{c}_1\right)+\hat{k}\left(c_2-b_2{c}_1\right)=c_1\hat{i}+c_2\hat{j}+c_3\hat{k}-3\hat{i}-\hat{j}+\hat{k}$

$\Rightarrow \overrightarrow{b}\times \overrightarrow{c}=\overrightarrow{c}-\overrightarrow{a} ....\left(\mathrm{i}\right)$

$\Rightarrow \left(\overrightarrow{b}\times \overrightarrow{c}\right)·\overrightarrow{b}=\overrightarrow{c}·\overrightarrow{b}-\overrightarrow{a}·\overrightarrow{b}$

$\Rightarrow \overrightarrow{b}·\overrightarrow{c}=0$ $\left\{\text{given }\overrightarrow{a}\text{.}\overrightarrow{b}\text{=0}\right\}$

Again from (i)

$\left(\overrightarrow{c}\times \overrightarrow{b}\right)·\overrightarrow{c}=\overrightarrow{a}·\overrightarrow{c}-{\left|\overrightarrow{c}\right|}^2=0$

$\Rightarrow {\left|\overrightarrow{c}\right|}^2=\left|\overrightarrow{a}\right|\left|\overrightarrow{c}\right|\cos \theta$, where $\theta ={\overrightarrow{a}}^{\wedge }\overrightarrow{c}$

$\Rightarrow \left|\overrightarrow{c}\right|\ge \left|\overrightarrow{a}\right|\Rightarrow \left|\overrightarrow{c}\right|\ge \sqrt{11}$

Given that $\overrightarrow{a}·\overrightarrow{b}=0\Rightarrow b_2-b_3+3=0$

$\Rightarrow b_3-b_2=3$

Also $b_2-b_3>0$

Now $\left|\overrightarrow{b}\right|=1+b_2^2+b_3^2$

$=1+{\left(b_3-b_2\right)}^2+2b_2{b}_3$

$=10+2b_2{b}_3$

$\Rightarrow {\left|\overrightarrow{b}\right|}^2>10\Rightarrow \left|\overrightarrow{b}\right|>\sqrt{10}$