Let $y=y(x)$ be the solution of the differential equation

$$ x \frac{d y}{d x}-\sin 2 y=x^3\left(2-x^3\right) \cos ^2 y, x \neq 0 . $$

If $y(2)=0$, then $\tan (y(1))$ is equal to

If $\frac{\tan (\mathrm{A}-\mathrm{B})}{\tan \mathrm{A}}+\frac{\sin ^2 \mathrm{C}}{\sin ^2 \mathrm{~A}}=1, \mathrm{~A}, \mathrm{~B}, \mathrm{C} \in\left(0, \frac{\pi}{2}\right)$, then

If the distances of the point $(1,2, a)$ from the line $\frac{x-1}{1}=\frac{y}{2}=\frac{z-1}{1}$ along the lines $\mathrm{L}_1: \frac{x-1}{3}=\frac{y-2}{4}=\frac{z-a}{b}$ and $\mathrm{L}_2: \frac{x-1}{1}=\frac{y-2}{4}=\frac{z-a}{c}$ are equal, then $a+b+c$ is equal to

For three unit vectors $\vec{a}, \vec{b}, \vec{c}$ satisfying

$$ |\vec{a}-\vec{b}|^2+|\vec{b}-\vec{c}|^2+|\vec{c}-\vec{a}|^2=9 \text { and }|2 \vec{a}+k \vec{b}+k \vec{c}|=3 \text {, } $$

the positive value of k is :