$${P_1} = I = \left[ {\matrix{ 1 & 0 & 0 \cr 0 & 1 & 0 \cr 0 & 0 & 1 \cr } } \right],\,{P_2} = \left[ {\matrix{ 1 & 0 & 0 \cr 0 & 0 & 1 \cr 0 & 1 & 0 \cr } } \right],\,{P_3} = \left[ {\matrix{ 0 & 1 & 0 \cr 1 & 0 & 0 \cr 0 & 0 & 1 \cr } } \right],\,{P_4} = \left[ {\matrix{ 0 & 1 & 0 \cr 0 & 0 & 1 \cr 1 & 0 & 0 \cr } } \right],\,{P_5} = \left[ {\matrix{ 0 & 0 & 1 \cr 1 & 0 & 0 \cr 0 & 1 & 0 \cr } } \right],\,{P_6} = \left[ {\matrix{ 0 & 0 & 1 \cr 0 & 1 & 0 \cr 1 & 0 & 0 \cr } } \right]$$ and $$X = \sum\limits_{k = 1}^6 {{P_k}} \left[ {\matrix{ 2 & 1 & 3 \cr 1 & 0 & 2 \cr 3 & 2 & 1 \cr } } \right]P_k^T$$

where $$P_k^T$$ denotes the transpose of the matrix P_k. Then which of the following option is/are correct?

Let $$M = \left[ {\matrix{ 0 & 1 & a \cr 1 & 2 & 3 \cr 3 & b & 1 \cr } } \right]$$ and

adj $$M = \left[ {\matrix{ { - 1} & 1 & { - 1} \cr 8 & { - 6} & 2 \cr { - 5} & 3 & { - 1} \cr } } \right]$$

where a and b are real numbers. Which of the following options is/are correct?

Let S be the set of all column matrices $$\left[ {\matrix{ {{b_1}} \cr {{b_2}} \cr {{b_3}} \cr } } \right]$$ such that $${b_1},{b_2},{b_3} \in R$$ and the system of equations (in real variables)

$$\eqalign{ & - x + 2y + 5z = {b_1} \cr & 2x - 4y + 3z = {b_2} \cr & x - 2y + 2z = {b_3} \cr} $$

has at least one solution. Then, which of the following system(s) (in real variables) has (have) at least one solution for each $$\left[ {\matrix{ {{b_1}} \cr {{b_2}} \cr {{b_3}} \cr } } \right]$$$$ \in $$S?

Which of the following is(are) NOT the square of a 3 $$ \times $$ 3 matrix with real entries?