Let $\mathrm{y}=\mathrm{y}(\mathrm{x})$ be the solution of the differential equation $\left(x y-5 x^2 \sqrt{1+x^2}\right) d x+\left(1+x^2\right) d y=0, y(0)=0$. Then $y(\sqrt{3})$ is equal to

Let in a $\triangle A B C$, the length of the side $A C$ be 6 , the vertex $B$ be $(1,2,3)$ and the vertices $A, C$ lie on the line $\frac{x-6}{3}=\frac{y-7}{2}=\frac{z-7}{-2}$. Then the area (in sq. units) of $\triangle A B C$ is:

Let the line passing through the points $(-1,2,1)$ and parallel to the line $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z}{4}$ intersect the line $\frac{x+2}{3}=\frac{y-3}{2}=\frac{z-4}{1}$ at the point $P$. Then the distance of $P$ from the point $Q(4,-5,1)$ is

$A$ and $B$ alternately throw a pair of dice. A wins if he throws a sum of 5 before $B$ throws a sum of 8 , and $B$ wins if he throws a sum of 8 before $A$ throws a sum of 5 . The probability, that A wins if A makes the first throw, is