Given,

$$\left[ {\matrix{ 0 & { - {c_3}} & {{c_2}} \cr {{c_3}} & 0 & { - {c_1}} \cr { - {c_2}} & {{c_1}} & 0 \cr } } \right]\left[ {\matrix{ 1 \cr {{b_2}} \cr {{b_3}} \cr } } \right] = \left[ {\matrix{ {3 - {c_1}} \cr {1 - {c_2}} \cr { - 1 - {c_3}} \cr } } \right]$$

$$\therefore$$ $$ - {b_2}{c_3} + {b_3}{c_2} = 3 - {c_1}$$ ..... (1)

$${c_3} - {b_3}{c_1} = 1 - {c_2}$$ ...... (2)

$$ - {c_2} + {b_2}{c_1} = - 1 - {c_3}$$ ..... (3)

Adding (1), (2) and (3), we get

$$({b_2} - {b_3}){c_1} + ({b_3} - 1){c_2} + (1 - {b_2}){c_3}$$

$$ = 3 - ({c_1} + {c_2} + {c_3})$$

$$ \Rightarrow ({c_2}{b_3} - {c_3}{b_2}) - ({c_1}{b_3} - {c_3}) + ({c_1}{b_2} - {c_2})$$

$$ = 3 - ({c_1} + {c_2} + {c_3})$$ ..... (4)

Also given,

$$\overrightarrow a = 3\widehat i + \widehat j - \widehat k$$

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

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

Now,

$$\overrightarrow c \times \overrightarrow b $$

$$ = \left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr {{c_1}} & {{c_2}} & {{c_3}} \cr 1 & {{b_2}} & {{b_3}} \cr } } \right|$$

$$ = ({c_2}{b_3} - {c_3}{b_2})\widehat i - \widehat j({c_1}{b_3} - {c_3}) + \widehat k({c_1}{b_2} - {c_2})$$

And

$$\overrightarrow a - \overrightarrow c = (3 - {c_1})\widehat i + (1 - {c_2})\widehat j + ( - 1 - {c_3})\widehat k$$

Comparing value of $$\overrightarrow c \times \overrightarrow b $$ and $$\overrightarrow a - \overrightarrow c $$ with equation (4), we get

$$\overrightarrow c \times \overrightarrow b = \overrightarrow a - \overrightarrow c $$ ..... (5)

Multiplying both side with $$\overrightarrow b $$, we get

$$\overrightarrow b \,.\,(\overrightarrow c \times \overrightarrow b ) = \overrightarrow a \,.\,\overrightarrow b - \overrightarrow b \,.\,\overrightarrow c $$

$$ \Rightarrow 0 = 0 - \overrightarrow b \,.\,\overrightarrow c $$ [Given $$\overrightarrow a \,.\,\overrightarrow b = 0$$]

$$ \Rightarrow \overrightarrow b \,.\,\overrightarrow c = 0$$

$$\therefore$$ (B) is correct.

Now multiplying both side of equation (5) with $$\overrightarrow c $$, we get

$$\overrightarrow c \,.\,(\overrightarrow c \times \overrightarrow b ) = \overrightarrow a \,.\,\overrightarrow c - |\overrightarrow c {|^2}$$

$$ \Rightarrow 0 = \overrightarrow a \,.\,\overrightarrow c - |\overrightarrow c {|^2}$$

$$ \Rightarrow \overrightarrow a \,.\,\overrightarrow c = |\overrightarrow c {|^2}$$

Now given $$\overrightarrow c = {c_1}\widehat i + {c_2}\widehat j + {c_3}\widehat k$$

$$\therefore$$ $$|\overrightarrow c {|^2} = \sqrt {c_1^2 + c_2^2 + c_3^2} $$

$$ \Rightarrow |\overrightarrow c {|^2} = c_1^2 + c_2^2 + c_3^2$$

$$\therefore$$ $$|\overrightarrow c {|^2} = 0$$ when $${c_1} = 0$$, $${c_2} = 0$$ and $${c_3} = 0$$

Now if we put $${c_1} = {c_2} = {c_3} = 0$$ in matrix we get,

$$\left[ {\matrix{ 0 & 0 & 0 \cr 0 & 0 & 0 \cr 0 & 0 & 0 \cr } } \right]\left[ {\matrix{ 1 \cr {{b_2}} \cr {{b_3}} \cr } } \right] = \left[ {\matrix{ {3 - {c_1}} \cr {1 - {c_2}} \cr { - 1 - {c_3}} \cr } } \right]$$

$$ \Rightarrow \left[ {\matrix{ 0 \cr 0 \cr 0 \cr } } \right] = \left[ {\matrix{ 3 \cr 1 \cr { - 1} \cr } } \right]$$

$$\therefore$$ Matrix does not satisfy the value of $${c_1} = {c_2} = {c_3} = 0$$.

$$\therefore$$ $${c_1},{c_2},{c_3}$$ can't be zero.

$$\therefore$$ $$|\overrightarrow c {|^2} \ne 0$$

$$ \Rightarrow \overrightarrow a \,.\,\overrightarrow c \ne 0$$

$$\therefore$$ Option (A) is wrong.

As $$\overrightarrow a \,.\,\overrightarrow c = |\overrightarrow c {|^2}$$

$$ \Rightarrow |\overrightarrow a ||\overrightarrow c |\cos \theta = |\overrightarrow c {|^2}$$ [$$\theta$$ = angle between $$\overrightarrow a $$ and $$\overrightarrow c $$]

$$ \Rightarrow |\overrightarrow a |\cos \theta = |\overrightarrow c |$$

$$ \Rightarrow (\sqrt {9 + 1 + 1} )\cos \theta = |\overrightarrow c |$$

$$ \Rightarrow |\overrightarrow c | = \sqrt {11} \cos \theta $$

$$ \Rightarrow |\overrightarrow c | \le \sqrt {11} $$ [as $$ - 1 \le \cos \theta \le 1$$]

$$\therefore$$ Option (D) is correct.

Given, $$\overrightarrow a \,.\,\overrightarrow b = 0$$

$$ \Rightarrow (3\widehat i + \widehat j - \widehat k)\,.\,(\widehat i + {b_2}\widehat j + {b_3}\widehat k) = 0$$

$$ \Rightarrow 3 + {b_2} - {b_3} = 0$$

$$ \Rightarrow {b_3} = 3 + {b_2}$$

Now,

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

$$ \Rightarrow |\overrightarrow b | = \sqrt {1 + b_2^2 + b_3^2} $$

$$ \Rightarrow |\overrightarrow b {|^2} = 1 + b_2^2 + {(3 + {b_2})^2}$$

$$ \Rightarrow |\overrightarrow b {|^2} = 1 + b_2^2 + 9 + 6{b_2} + b_2^2$$

$$ \Rightarrow |\overrightarrow b {|^2} = 10 + 6{b_2} + 2b_2^2$$

$$ \Rightarrow |\overrightarrow b {|^2} = 10 + 2{b_2}(3 + {b_2})$$

$$ \Rightarrow |\overrightarrow b {|^2} = 10 + 2{b_2}{b_3}$$

$$ \Rightarrow |\overrightarrow b {|^2} > 10$$ [as $${b_2}{b_3} > 0$$ then $$2{b_2}{b_3} > 0$$]

$$ \Rightarrow |\overrightarrow b | > \sqrt {10} $$

$$\therefore$$ Option (C) is correct.