Let $D$ be the foot of the perpendicular drawn from the point $A(2,0,3)$ to the line joining the points $B(0,4,1)$ and $C(-2,0,4)$. Then, the ratio in which $D$ divides $B C$ is

Let $6 x-3 y+2 z-6=0$ be the given plane. If $a, b$ and $c$ are the intercepts made by the plane on $X, Y$ and $Z$-axes, respectively; $l, m$ and $n$ are the direction cosines of a normal drawn to the plane and $p$ is the perpendicular distance from the origin to the plane, then $|a l+b m+c n|=$

$$ \mathop {\lim }\limits_{x \to 0} \frac{\tan ^2\left(\pi \sec ^4 x\right)}{\pi^2 x^4}= $$

$$\mathop {\lim }\limits_{x \to 0}\left(\frac{4!}{x^8}\left(1-\cos \frac{x^2}{3}-\cos \frac{x^2}{4}+\cos \frac{x^2}{3} \cos \frac{x^2}{4}\right)\right)= $$