Let $\overrightarrow{\mathrm{a}}=4 \hat{i}-\hat{j}+3 \hat{k}, \overrightarrow{\mathrm{~b}}=10 \hat{i}+2 \hat{j}-\hat{k}$ and a vector $\overrightarrow{\mathrm{c}}$ be such that $2(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}})+3(\overrightarrow{\mathrm{~b}} \times \overrightarrow{\mathrm{c}})=\overrightarrow{0}$.

If $\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{c}}=15$, then $\overrightarrow{\mathrm{c}} \cdot(\hat{i}+\hat{j}-3 \hat{k})$ is equal to :

Let the foot of perpendicular from the point $(\lambda, 2,3)$ on the line $\frac{x-4}{1}=\frac{y-9}{2}=\frac{z-5}{1}$ be the point ( $1, \mu, 2$ ). Then the distance between the lines $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z+4}{6}$ and $\frac{x-\lambda}{2}=\frac{y-\mu}{3}=\frac{z+5}{6}$ is equal to :

The value of the integral $\int\limits_0^2 \frac{\sqrt{x\left(x^2+x+1\right)}}{(\sqrt{x+1})\left(\sqrt{x^4+x^2+1}\right)} \mathrm{d} x$ is equal to:

Let $y=y(x)$ be the solution of the differential equation $x \sqrt{1-x^2} d y+\left(y \sqrt{1-x^2}-x \cos ^{-1} x\right) d x=0, x \in(0,1), \lim _{x \rightarrow 1^{-}} y(x)=1$. Then $y\left(\frac{1}{2}\right)$ equals :