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

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

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

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

be three vectors such that $$b_2 b_3 > 0$$, $$\vec{a} \cdot \vec{b} = 0$$ and $$\begin{pmatrix} 0 & -c_3 & c_2 \\ c_3 & 0 & -c_1 \\ -c_2 & c_1 & 0 \end{pmatrix} \begin{pmatrix} 1 \\ b_2 \\ b_3 \end{pmatrix} = \begin{pmatrix} 3 - c_1 \\ 1 - c_2 \\ -1 - c_3 \end{pmatrix}$$.

Then, which of the following is/are TRUE?

We are given

$$\vec a = 3\hat i+\hat j-\hat k,\qquad \vec b = \hat i+b_2\hat j+b_3\hat k,\qquad \vec c = c_1\hat i+c_2\hat j+c_3\hat k,\qquad b_2b_3\gt 0.$$ The conditions are

1. $$\vec a\cdot\vec b = 0,$$

2. $$\begin{pmatrix} 0 & -c_3 & c_2 \\ c_3 & 0 & -c_1 \\ -c_2 & c_1 & 0 \end{pmatrix} \begin{pmatrix} 1 \\ b_2 \\ b_3 \end{pmatrix} = \begin{pmatrix} 3-c_1 \\ 1-c_2 \\ -1-c_3 \end{pmatrix}.$$ All assertions have to be checked one by one.

Step 1: Using $$\vec a\cdot\vec b=0$$
$$\vec a\cdot\vec b = 3(1)+1(b_2)+(-1)(b_3)=0 \;\Longrightarrow\; 3+b_2-b_3=0 \;\Longrightarrow\; b_3 = 3+b_2.$$

The extra restriction $$b_2b_3\gt 0$$ now reads

Either $$b_2\gt 0\quad(\text{then }b_3=3+b_2\gt 3),$$
or $$b_2\lt -3\quad(\text{then }b_3=3+b_2\lt 0).$$

Step 2: Translating the matrix equation into three scalar equations

The given matrix equation is equivalent to

$$\begin{cases} -b_2c_3 + (3+b_2)c_2 + c_1 = 3 & -(A)\\[2pt] \;c_3 - (3+b_2)c_1 + c_2 = 1 & -(B)\\[2pt] \;c_3 + b_2c_1 - c_2 = -1 & -(C) \end{cases}$$

Step 3: Solving (B) and (C)

Adding (B) and (C): $$2c_3 - 3c_1 = 0 \;\Longrightarrow\; c_3=\dfrac32\,c_1.$$
Subtracting (C) from (B): $$2c_2-(3+2b_2)c_1=2 \;\Longrightarrow\; c_2 = 1+\dfrac{3+2b_2}{2}\,c_1.$$ Put $$k=\dfrac{3+2b_2}{2}\quad\bigl(\text{so }c_2=1+k\,c_1\bigr).$$

Step 4: Using equation (A) to determine $$c_1$$

Insert the relations for $$c_2$$ and $$c_3$$ in (A):

$$-b_2\!\left(\tfrac32c_1\right) +(3+b_2)(1+k\,c_1)+c_1 = 3.$$

Simplifying, $$\bigl(b_2^2+3b_2+\tfrac{11}{2}\bigr)c_1 = -b_2.$$ Define $$D=b_2^2+3b_2+\tfrac{11}{2}>0,$$ then

$$c_1 = -\dfrac{\,b_2\,}{D},\qquad c_2 = 1+k\,c_1,\qquad c_3 = \dfrac32c_1.$$

Step 5: Dot-product $$\vec b\cdot\vec c$$

$$\vec b\cdot\vec c = 1\cdot c_1 + b_2c_2 + (3+b_2)c_3.$$ Substitute $$c_2=1+k\,c_1,\; c_3=\tfrac32c_1$$ to get

$$\vec b\cdot\vec c = b_2 + \bigl(1+b_2k+\tfrac32(3+b_2)\bigr)c_1.$$ But the coefficient $$1+b_2k+\tfrac32(3+b_2)=D$$ is exactly the same $$D$$ introduced above. Hence $$\vec b\cdot\vec c = b_2 - D\,\dfrac{b_2}{D}=0.$$ So statement B ( $$\vec b\cdot\vec c=0$$ ) is TRUE.

Step 6: Magnitude of $$\vec b$$

$$|\vec b|^2 = 1+b_2^2+b_3^2 = 1+b_2^2+(3+b_2)^2 = 2b_2^2+6b_2+10 = 2b_2(b_2+3)+10.$$ Because $$b_2$$ and $$(b_2+3)$$ have the same sign (see Step 1), the product $$2b_2(b_2+3)$$ is \emph{strictly positive}. Therefore

$$|\vec b|^2 \gt 10 \;\Longrightarrow\; |\vec b| \gt \sqrt{10}.$$ Hence statement C is TRUE.

Step 7: Magnitude of $$\vec c$$

Using the expressions found for $$c_1,c_2,c_3$$:

$$\begin{aligned} |\vec c|^2 &= c_1^2+c_2^2+c_3^2 \\[2pt] &= c_1^2+\bigl(1+k\,c_1\bigr)^2+\bigl(\tfrac32c_1\bigr)^2 \\[2pt] &= 1-\dfrac{2k\,b_2}{D}+\dfrac{(3.25+k^2)b_2^2}{D^2}\\[2pt] &= \dfrac{11/2}{\,D\,}. \end{aligned}$$ (Every algebraic term collapses to the simple fraction shown.)

Because $$D=b_2^2+3b_2+\tfrac{11}{2}\ge \tfrac{13}{4} \;(\text{always})$$ we get

$$|\vec c|^2 \le \dfrac{11}{2}\;\Big/\;\dfrac{13}{4} = \dfrac{22}{13} \lt 11.$$ Thus $$|\vec c|\le \sqrt{11},$$ so statement D is TRUE.

Step 8: Checking $$\vec a\cdot\vec c$$

$$\vec a\cdot\vec c = 3c_1+c_2-c_3 = 1+\bigl(3+b_2\bigr)c_1 = 1-\dfrac{b_2(3+b_2)}{D} = \dfrac{11/2}{\,D\,},$$ the same positive quantity obtained in Step 7. Hence $$\vec a\cdot\vec c\neq 0,$$ so statement A is FALSE.

Conclusion
The correct statements are
Option B ( $$\vec b\cdot\vec c=0$$ ),
Option C ( $$|\vec b|\gt\sqrt{10}$$ ) and
Option D ( $$|\vec c|\le\sqrt{11}$$ ).

Answer: Option A is false; Options B, C and D are TRUE.