Suppose that $$\overrightarrow p ,\overrightarrow q $$ and $$\overrightarrow r $$ are three non-coplanar vectors in $${R^3}$$. Let the components of a vector $$\overrightarrow s $$ along $$\overrightarrow p ,$$ $$\overrightarrow q $$ and $$\overrightarrow r $$ be $$4, 3$$ and $$5,$$ respectively. If the components of this vector $$\overrightarrow s $$ along $$\left( { - \overrightarrow p + \overrightarrow q + \overrightarrow r } \right),\left( {\overrightarrow p - \overrightarrow q + \overrightarrow r } \right)$$ and $$\left( { - \overrightarrow p - \overrightarrow q + \overrightarrow r } \right)$$ are $$x, y$$ and $$z,$$ respectively, then the value of $$2x+y+z$$ is

Let $${n_1}$$ and $${n_2}$$ be the number of red and black balls, respectively, in box $${\rm I}$$. Let $${n_3}$$ and $${n_4}$$ be the number of red and black balls, respectively, in box $${\rm I}{\rm I}.$$

A ball is drawn at random from box $${\rm I}$$ and transferred to box $${\rm I}$$$${\rm I}.$$ If the probability of drawing a red ball from box $${\rm I},$$ after this transfer, is $${1 \over 3},$$ then the correct option(s) with the possible values of $${n_1}$$ and $${n_2}$$ is(are)

Let $${n_1}$$ and $${n_2}$$ be the number of red and black balls, respectively, in box $${\rm I}$$. Let $${n_3}$$ and $${n_4}$$ be the number of red and black balls, respectively, in box $${\rm I}{\rm I}.$$

One of the two boxes, box $${\rm I}$$ and box $${\rm I}{\rm I},$$ was selected at random and a ball was drawn randomly out of this box. The ball was found to be red. If the probability that this red ball was drawn from box $${\rm I}{\rm I}$$ is $${1 \over 3},$$ then the correct option(s) with the possible values of $${n_1}$$ $${n_2},$$ $${n_3}$$ and $${n_4}$$ is (are)

If $$\alpha = \int\limits_0^1 {\left( {{e^{9x + 3{{\tan }^{ - 1}}x}}} \right)\left( {{{12 + 9{x^2}} \over {1 + {x^2}}}} \right)} dx$$ where $${\tan ^{ - 1}}x$$ takes only principal values, then the value of $$\left( {{{\log }_e}\left| {1 + \alpha } \right| - {{3\pi } \over 4}} \right)$$ is