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

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

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

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

be three vectors such that $b_2{b}_3>0$, $\overrightarrow{a}\cdot \overrightarrow{b}=0$ and

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

Then, which of the following is/are TRUE?

The correct options are:
1. $\overrightarrow{b}\cdot \overrightarrow{c}=0$
2. $\left|\overrightarrow{b}\right|>\sqrt{10}$
3. $\left|\overrightarrow{c}\right|\le \sqrt{11}$

Step-by-Step Explanation:

Step 1: Express given conditions using vector operations

We are given the vectors:
$\overrightarrow{a}=3\hat{i}+\hat{j}-\hat{k}$
$\overrightarrow{b}=\hat{i}+b_2\hat{j}+b_3\hat{k}$
$\overrightarrow{c}=c_1\hat{i}+c_2\hat{j}+c_3\hat{k}$

We are given that $\overrightarrow{a}\cdot \overrightarrow{b}=0$:
$\left(3\right)\left(1\right)+\left(1\right)\left(b_2\right)+(-1)\left(b_3\right)=0$
$\Rightarrow 3+b_2-b_3=0$
$\Rightarrow b_3-b_2=3$

Step 2: Simplify the matrix equation

Notice that the matrix equation:
$\left[\begin{array}{ccc}0& -c_3& c_2\\ c_3& 0& -c_1\\ -c_2& c_1& 0\end{array}\right]\left[\begin{array}{c}1\\ b_2\\ b_3\end{array}\right]=\left[\begin{array}{c}3-c_1\\ 1-c_2\\ -1-c_3\end{array}\right]$
represents the vector cross product $\overrightarrow{c}\times \overrightarrow{b}$ on the left side, and $\overrightarrow{a}-\overrightarrow{c}$ on the right side.
Hence, we get:
$\overrightarrow{c}\times \overrightarrow{b}=\overrightarrow{a}-\overrightarrow{c}$

Step 3: Analyze dot products

Take the dot product of both sides with $\overrightarrow{b}$:
$\overrightarrow{b}\cdot (\overrightarrow{c}\times \overrightarrow{b})=\overrightarrow{b}\cdot (\overrightarrow{a}-\overrightarrow{c})$
Since $\overrightarrow{b}\cdot (\overrightarrow{c}\times \overrightarrow{b})=0$, we have:
$0=\overrightarrow{b}\cdot \overrightarrow{a}-\overrightarrow{b}\cdot \overrightarrow{c}$
Given , this simplifies to:
$\overrightarrow{b}\cdot \overrightarrow{c}=0$
This proves that $\overrightarrow{b}\cdot \overrightarrow{c}=0$ is TRUE.

Now, take the dot product of both sides with $\overrightarrow{c}$:
$\overrightarrow{c}\cdot (\overrightarrow{c}\times \overrightarrow{b})=\overrightarrow{c}\cdot (\overrightarrow{a}-\overrightarrow{c})$
$0=\overrightarrow{a}\cdot \overrightarrow{c}-|\overrightarrow{c}{|}^2$
$\Rightarrow \overrightarrow{a}\cdot \overrightarrow{c}=|\overrightarrow{c}{|}^2$

Step 4: Determine magnitude of $\overrightarrow{b}$

We are given $b_3-b_2=3\Rightarrow b_3=b_2+3$ and $b_2{b}_3>0$.
Since $b_2(b_2+3)>0$, either $b_2>0$ or $b_2<-3$.
Now consider:
$|\overrightarrow{b}{|}^2=1+b_2^2+b_3^2=1+b_2^2+(b_2+3{)}^2=1+2b_2^2+6b_2+9=10+2b_2(b_2+3)$
Since $b_2(b_2+3)>0$, we have:
$|\overrightarrow{b}{|}^2>10\Rightarrow |\overrightarrow{b}|>\sqrt{10}$
This proves that $\left|\overrightarrow{b}\right|>\sqrt{10}$ is TRUE.

Step 5: Determine magnitude of $\overrightarrow{c}$

Taking the magnitude squared of both sides of $\overrightarrow{c}\times \overrightarrow{b}=\overrightarrow{a}-\overrightarrow{c}$:
$|\overrightarrow{c}\times \overrightarrow{b}{|}^2=|\overrightarrow{a}-\overrightarrow{c}{|}^2$
Since $\overrightarrow{b}\cdot \overrightarrow{c}=0$, the left side is $\left|\overrightarrow{c}{|}^2\right|\overrightarrow{b}{|}^2$.
The right side is:
$|\overrightarrow{a}-\overrightarrow{c}{|}^2=|\overrightarrow{a}{|}^2+|\overrightarrow{c}{|}^2-2(\overrightarrow{a}\cdot \overrightarrow{c})$
Substituting $\overrightarrow{a}\cdot \overrightarrow{c}=|\overrightarrow{c}{|}^2$ into this expression gives:
$|\overrightarrow{a}-\overrightarrow{c}{|}^2=|\overrightarrow{a}{|}^2-|\overrightarrow{c}{|}^2$
Equating both sides:
$\left|\overrightarrow{c}{|}^2\right|\overrightarrow{b}{|}^2=|\overrightarrow{a}{|}^2-|\overrightarrow{c}{|}^2$
$\Rightarrow \left|\overrightarrow{c}{|}^2\right(|\overrightarrow{b}{|}^2+1)=|\overrightarrow{a}{|}^2$

Calculating $|\overrightarrow{a}{|}^2$:
$|\overrightarrow{a}{|}^2=3^2+1^2+(-1{)}^2=11$

Thus:
$|\overrightarrow{c}{|}^2=\frac{11}{|\overrightarrow{b}{|}^2+1}$
Since $|\overrightarrow{b}{|}^2>10$, we have $|\overrightarrow{b}{|}^2+1>11$, so:
$|\overrightarrow{c}{|}^2<\frac{11}{11}=1\Rightarrow |\overrightarrow{c}|<1$
Since $\left|\overrightarrow{c}\right|<1$, it is strictly less than $\sqrt{11}$, which means $\left|\overrightarrow{c}\right|\le \sqrt{11}$ is TRUE.