$$P = \left[ {\matrix{ 3 & { - 1} & { - 2} \cr 2 & 0 & \alpha \cr 3 & { - 5} & 0 \cr } } \right]$$

Now, $$\left| P \right| = 3(5\alpha ) + 1( - 3\alpha ) - 2( - 10)$$

$$ = 12\alpha + 20$$ ....... (i)

$$\therefore$$ $$(P) = {\left[ {\matrix{ {5\alpha } & {2\alpha } & { - 10} \cr { - 10} & 6 & {12} \cr { - \alpha } & { - (3\alpha + 4)} & 2 \cr } } \right]^T}$$

$$ = \left[ {\matrix{ {5\alpha } & { - 10} & { - \alpha } \cr {2\alpha } & 6 & { - 3\alpha - 4} \cr { - 10} & {12} & 2 \cr } } \right]$$ ...... (ii)

As, $$PQ = kI$$

$$ \Rightarrow \left| P \right|\left| Q \right| = \left| {kI} \right|$$

$$ \Rightarrow \left| P \right|\left| Q \right| = {k^3}$$

$$ \Rightarrow \left| P \right|\left( {{{{k^2}} \over 2}} \right) = {k^3}$$ [given, $$\left| Q \right| = {{{k^2}} \over 2}$$]

$$ \Rightarrow \left| P \right| = 2k$$ ......(iii)

$$\because$$ $$PQ = ki$$

$$\therefore$$ $$Q = k{p^{ - 1}}I$$

$$ = k\,.\,{{adj\,P} \over {\left| P \right|}} = {{k(adj\,P)} \over {2k}}$$ [from Eq. (iii)]

$$ = {{adj\,P} \over 2}$$

$$ = {1 \over 2}\left[ {\matrix{ {5\alpha } & { - 10} & { - \alpha } \cr {2\alpha } & 6 & { - 3\alpha - 4} \cr { - 10} & {12} & 2 \cr } } \right]$$

$$\therefore$$ $${q_{23}} = {{ - 3\alpha - 4} \over 2}$$ [given, $${q_{23}} = - {k \over 8}$$]

$$ \Rightarrow - {{(3\alpha + 4)} \over 2} = - {k \over 8}$$

$$ \Rightarrow (3\alpha + 4) \times 4 = k$$

$$ \Rightarrow 12\alpha + 16 = k$$ ...... (iv)

From Eq. (iii), $$\left| P \right| = 2k$$

$$ \Rightarrow 12\alpha + 20 = 2k$$ [from Eq. (i)] ...... (v)

On solving Eqs. (iv) and (v), we get

$$\alpha$$ = $$-$$1 and k = 4 ....... (vi)

$$\therefore$$ $$4\alpha - k + 8 = - 4 - 4 + 8 = 0$$

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

Now, $$\left| {P\,adj(Q)} \right| = \left| P \right|\left| {adj\,Q} \right|$$

$$ = 2k{\left( {{{{k^2}} \over 2}} \right)^2} = {{{k^5}} \over 2} = {{{2^{10}}} \over 2} = {2^9}$$

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

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