If $\sin \left(\tan ^{-1}(x \sqrt{2})\right)=\cot \left(\sin ^{-1} \sqrt{1-x^2}\right), x \in(0,1)$, then the value of $x$ is:

The shortest distance between the lines $\frac{x-4}{1}=\frac{y-3}{2}=\frac{z-2}{-3}$ and $\frac{x+2}{2}=\frac{y-6}{4}=\frac{z-5}{-5}$ is:

Let $\overrightarrow{\mathrm{a}}=2 \hat{i}+3 \hat{j}+3 \hat{k}$ and $\overrightarrow{\mathrm{b}}=6 \hat{i}+3 \hat{j}+3 \hat{k}$. Then the square of the area of the triangle with adjacent sides determined by the vectors $(2 \vec{a}+3 \vec{b})$ and $(\vec{a}-\vec{b})$ is :

Let $\mathop {\lim }\limits_{x \to 2} \frac{(\tan (x-2))\left(\mathrm{r} x^2+(\mathrm{p}-2) x-2 \mathrm{p}\right)}{(x-2)^2}=5$ for some $\mathrm{r}, \mathrm{p} \in \boldsymbol{R}$. If the set of all possible values of q , such that the roots of the equation $\mathrm{r} x^2-\mathrm{p} x+\mathrm{q}=0$ lie in $(0,2)$, be the interval $(\alpha, \beta]$, then $4(\alpha+\beta)$ equals :