The correct statements about of the following reaction sequence is (are)

Cumene (C₉H₁₂) $$\mathrel{\mathop{\kern0pt\longrightarrow} \limits_{(ii)\,{H_3}{O^ + }}^{(i)\,{O_2}}} $$ P $$\mathrel{\mathop{\kern0pt\longrightarrow} \limits_{}^{CHC{l_3}/NaOH}} $$ Q (major) + R (minor)

Q $$\mathrel{\mathop{\kern0pt\longrightarrow} \limits_{PhC{H_2}Br}^{NaOH}} $$ S

Let $$S = \left\{ {x \in \left( { - \pi ,\pi } \right):x \ne 0, \pm {\pi \over 2}} \right\}.$$ The sum of all distinct solutions of the equation $$\sqrt 3 \,\sec x + \cos ec\,x + 2\left( {\tan x - \cot x} \right) = 0$$ in the set S is equal to

Consider a pyramid $$OPQRS$$ located in the first octant $$\left( {x \ge 0,y \ge 0,z \ge 0} \right)$$ with $$O$$ as origin, and $$OP$$ and $$OR$$ along the $$x$$-axis and the $$y$$-axis, respectively. The base $$OPQR$$ of the pyramid is a square with $$OP=3.$$ The point $$S$$ is directly above the mid-point, $$T$$ of diagonal $$OQ$$ such that $$TS=3.$$ Then

A computer producing factory has only two plants $${T_1}$$ and $${T_2}.$$ Plant $${T_1}$$ produces $$20$$% and plant $${T_2}$$ produces $$80$$% of the total computers produced. $$7$$% of computers produced in the factory turn out to be defective. It is known that $$P$$ (computer turns out to be defective given that it is produced in plant $${T_1}$$)
$$ = 10P$$ (computer turns out to be defective given that it is produced in plant $${T_2}$$),
where $$P(E)$$ denotes the probability of an event $$E$$. A computer produced in the factory is randomly selected and it does not turn out to be defective. Then the probability that it is produced in plant $${T_2}$$ is