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

$$ \because $$ $$P_1^T = {P_1},\,P_2^T = {P_2},\,P_3^T = {P_3},\,P_4^T = {P_5},\,P_5^T = {P_4}$$ and $$P_6^T = {P_6}$$

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

and $$ \because $$ Q^T = Q

Now,

$$X = ({P_1}QP_1^T) + ({P_2}QP_2^T) + ({P_3}QP_3^T) + ({P_4}QP_4^T) + ({P_5}QP_5^T) + ({P_6}QP_6^T)$$

So, $${X^T} = {({P_1}QP_1^T)^T} + {({P_2}QP_2^T)^T} + {({P_3}QP_3^T)^T} + {({P_4}QP_4^T)^T} + {({P_5}QP_5^T)^T} + {({P_6}QP_6^T)^T}$$

= $${P_1}QP_1^T$$ + $${P_2}QP_2^T$$ + $${P_3}QP_3^T$$ + $${P_4}QP_4^T$$ + $${P_5}QP_5^T$$ + $${P_6}QP_6^T$$

[$$ \because $$ (ABC)^T = C^TB^TA^T and (A^T)^T = A and Q^T = Q]

$$ \Rightarrow $$ X^T = X

$$ \Rightarrow $$ X is a symmetric matrix.

The sum of diagonal entries of X = Tr(X)

$$ = \sum\limits_{i = 1}^6 {{T_r}({P_i}QP_i^T) = \sum\limits_{i = 1}^6 {{T_r}(QP_i^T{P_i})} } $$ [$$ \because $$ T_r(ABC) = T_r(BCA)]

$$ = \sum\limits_{i = 1}^6 {{T_r}(QI)} $$

[$$ \because $$ P_i's are orthogonal matrices]

$$ = \sum\limits_{i = 1}^6 {{T_r}(Q) = 6{T_r}(Q) = 6 \times 3 = 18} $$

Now, Let $$R = \left[ {\matrix{ 1 \cr 1 \cr 1 \cr } } \right]$$, then

$$XR = \sum\limits_{K = 1}^6 {({P_K}} QP_K^T)R = \sum\limits_{K = 1}^6 {({P_K}QP_K^TR)} $$

$$ = \sum\limits_{K = 1}^6 {({P_K}QR)} $$ [$$ \because $$ $$P_K^T$$R = R]

$$ = \sum\limits_{K = 1}^6 {{P_K}\left[ {\matrix{ 6 \cr 3 \cr 6 \cr } } \right]} $$

$$ = \sum\limits_{K = 1}^6 {{P_K}\left[ {\matrix{ 6 \cr 3 \cr 6 \cr } } \right]} = \left[ {\matrix{ 2 & 2 & 2 \cr 2 & 2 & 2 \cr 2 & 2 & 2 \cr } } \right]\left[ {\matrix{ 6 \cr 3 \cr 6 \cr } } \right]$$

$$ \Rightarrow XR = \left[ {\matrix{ {30} \cr {30} \cr {30} \cr } } \right] \Rightarrow XR = 30R$$

$$ \Rightarrow X\left[ {\matrix{ 1 \cr 1 \cr 1 \cr } } \right] = 30\left[ {\matrix{ 1 \cr 1 \cr 1 \cr } } \right]$$

$$ \Rightarrow $$ (X $$-$$ 30I) R = 0 $$ \Rightarrow $$ |X $$-$$ 30I| = 0

So, (X $$-$$ 30I) is not invertible and value of $$\alpha $$ = 30.

Hence, options (a), (b) and (d) are correct.