It is given, that matrices

$$P = \left[ {\matrix{ 1 & 1 & 1 \cr 0 & 2 & 2 \cr 0 & 0 & 3 \cr } } \right]$$, $$Q = \left[ {\matrix{ 2 & x & x \cr 0 & 4 & 0 \cr x & x & 6 \cr } } \right]$$

$$ \therefore $$ $${P^{ - 1}} = {{adj(P)} \over {|P|}}$$

as |P| = 6 and adj P = $${\left[ {\matrix{ 6 & 0 & 0 \cr { - 3} & 3 & 0 \cr 0 & { - 2} & 2 \cr } } \right]^T}$$

$$ \Rightarrow $$ $${P^{ - 1}} = {1 \over 6}\left[ {\matrix{ 6 & { - 3} & 0 \cr 0 & 3 & { - 2} \cr 0 & 0 & 2 \cr } } \right]$$

$$ \therefore $$ |R| = |PQP^$$-$$1| [$$ \because $$ R = PQP^$$-$$1 (given)]

$$ \Rightarrow $$ |R| = |P||Q||P^$$-$$1| = |Q| [$$ \because $$ |P| |P^$$-$$1| = |I| = 1]

$$ = \left[ {\matrix{ 2 & x & x \cr 0 & 4 & 0 \cr x & x & 6 \cr } } \right] = \left[ {\matrix{ 2 & x & x \cr 0 & 4 & 0 \cr x & x & 5 \cr } } \right] + \left[ {\matrix{ 2 & x & 0 \cr 0 & 4 & 0 \cr x & x & 1 \cr } } \right]$$

$$ = \left[ {\matrix{ 2 & x & x \cr 0 & 4 & 0 \cr x & x & 5 \cr } } \right] + 2(4 - 0) - x(0 - 0) + 0(0 - 4x)$$

$$ = \left[ {\matrix{ 2 & x & x \cr 0 & 4 & 0 \cr x & x & 5 \cr } } \right] + 8$$, for all x $$ \in $$ R

$$ \because $$ $$PQ = \left[ {\matrix{ 1 & 1 & 1 \cr 0 & 2 & 2 \cr 0 & 0 & 3 \cr } } \right]\left[ {\matrix{ 2 & x & x \cr 0 & 4 & 0 \cr x & x & 6 \cr } } \right] = \left[ {\matrix{ {2 + x} & {4 + 2x} & {x + 6} \cr {2x} & {2x + 8} & {12} \cr {3x} & {3x} & {18} \cr } } \right]$$

and $$QP = \left[ {\matrix{ 2 & x & x \cr 0 & 4 & 0 \cr x & x & 6 \cr } } \right]\left[ {\matrix{ 1 & 1 & 1 \cr 0 & 2 & 2 \cr 0 & 0 & 3 \cr } } \right] = \left[ {\matrix{ 2 & {2 + 2x} & {2 + 5x} \cr 0 & 8 & 8 \cr x & {3x} & {3x + 18} \cr } } \right]$$

There is no common value of 'x', for which each corresponding element of matrices PQ and QP is equal.

For x = 0, $$Q = \left[ {\matrix{ 2 & 0 & 0 \cr 0 & 4 & 0 \cr 0 & 0 & 6 \cr } } \right]$$

then, if $$R\left[ {\matrix{ 1 \cr a \cr b \cr } } \right] = 6\left[ {\matrix{ 1 \cr a \cr b \cr } } \right]$$

$$ \Rightarrow $$ $$PQ{P^{ - 1}}\left[ {\matrix{ 1 \cr a \cr b \cr } } \right] = 6\left[ {\matrix{ 1 \cr a \cr b \cr } } \right]$$ [$$ \because $$ R = PQP^$$-$$1]

$$ \Rightarrow {1 \over 6}\left[ {\matrix{ 1 & 1 & 1 \cr 0 & 2 & 2 \cr 0 & 0 & 3 \cr } } \right]\left[ {\matrix{ 2 & 0 & 0 \cr 0 & 4 & 0 \cr 0 & 0 & 6 \cr } } \right]\left[ {\matrix{ 6 & { - 3} & 0 \cr 0 & 3 & { - 2} \cr 0 & 0 & 2 \cr } } \right]\left[ {\matrix{ 1 \cr a \cr b \cr } } \right] = 6\left[ {\matrix{ 1 \cr a \cr b \cr } } \right]$$

$$ \Rightarrow {1 \over 6}\left[ {\matrix{ 2 & 4 & 6 \cr 0 & 8 & {12} \cr 0 & 0 & {18} \cr } } \right]\left[ {\matrix{ 6 & { - 3} & 0 \cr 0 & 3 & { - 2} \cr 0 & 0 & 2 \cr } } \right]\left[ {\matrix{ 1 \cr a \cr b \cr } } \right] = 6\left[ {\matrix{ 1 \cr a \cr b \cr } } \right]$$

$$ \Rightarrow \left[ {\matrix{ {12} & 6 & 4 \cr 0 & {24} & 8 \cr 0 & 0 & {36} \cr } } \right]\left[ {\matrix{ 1 \cr a \cr b \cr } } \right] = 36\left[ {\matrix{ 1 \cr a \cr b \cr } } \right]$$

$$ \Rightarrow \left[ {\matrix{ {12} & + & {6a} & + & {4b} \cr 0 & + & {24a} & + & {8b} \cr 0 & + & 0 & + & {36b} \cr } } \right] = \left[ {\matrix{ {36} \cr {36a} \cr {36b} \cr } } \right]$$

$$ \Rightarrow 6a + 4b = 24$$ and $$12a = 8b$$

$$ \Rightarrow 3a + 2b = 12$$ and $$3a = 2b$$

$$ \Rightarrow a = 2$$ and $$b = 3$$

So, a + b = 5.

Now, $$R\left[ {\matrix{ \alpha \cr \beta \cr \gamma \cr } } \right] = \left[ {\matrix{ 0 \cr 0 \cr 0 \cr } } \right]$$

and $$\alpha \widehat i + \beta \widehat j + \gamma \widehat k$$ is a unit vector,

so det (R) = 0

$$ \Rightarrow $$ det (Q) = 0 [$$ \because $$ R = PQP^$$-$$1 So, |R| = |Q|]

$$ \Rightarrow \left| {\matrix{ 2 & x & x \cr 0 & 4 & 0 \cr x & x & 6 \cr } } \right| = 0$$

$$ \Rightarrow 2(24 - 0) - x(0 - 0) + x(0 - 4x) = 0 \Rightarrow 48 - 4{x^2} = 0$$

$$ \Rightarrow {x^2} = 12 \Rightarrow x = \pm 2\sqrt 3 $$

So, for x = 1, there does not exist a unit vector $$\alpha \widehat i + \beta \widehat j + \gamma \widehat k$$, for which $$R\left[ {\matrix{ \alpha \cr \beta \cr \gamma \cr } } \right] = \left[ {\matrix{ 0 \cr 0 \cr 0 \cr } } \right]$$

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