The distance of the line $\frac{x-2}{2}=\frac{y-6}{3}=\frac{z-3}{4}$ from the point $(1,4,0)$ along the line $\frac{x}{1}=\frac{y-2}{2}=\frac{z+3}{3}$ is :

$\lim _{x \rightarrow \infty} \frac{\left(2 x^2-3 x+5\right)(3 x-1)^{\frac{x}{2}}}{\left(3 x^2+5 x+4\right) \sqrt{(3 x+2)^x}}$ is equal to :

Let $A=\left[a_{i j}\right]$ be a $3 \times 3$ matrix such that $A\left[\begin{array}{l}0 \\ 1 \\ 0\end{array}\right]=\left[\begin{array}{l}0 \\ 0 \\ 1\end{array}\right], A\left[\begin{array}{l}4 \\ 1 \\ 3\end{array}\right]=\left[\begin{array}{l}0 \\ 1 \\ 0\end{array}\right]$ and $A\left[\begin{array}{l}2 \\ 1 \\ 2\end{array}\right]=\left[\begin{array}{l}1 \\ 0 \\ 0\end{array}\right]$, then $a_{23}$ equals :

Let $\mathrm{A}=\{(x, y) \in \mathbf{R} \times \mathbf{R}:|x+y| \geqslant 3\}$ and $\mathrm{B}=\{(x, y) \in \mathbf{R} \times \mathbf{R}:|x|+|y| \leq 3\}$. If $\mathrm{C}=\{(x, y) \in \mathrm{A} \cap \mathrm{B}: x=0$ or $y=0\}$, then $\sum_{(x, y) \in \mathrm{C}}|x+y|$ is :