The shortest distance between the lines

$$ \begin{aligned} & \mathbf{r}=(3 \hat{\mathbf{i}}-5 \hat{\mathbf{j}}+2 \hat{\mathbf{k}})+t(4 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-\hat{\mathbf{k}}) \text { and } \\ & \mathbf{r}=(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-4 \hat{\mathbf{k}})+s(6 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-2 \hat{\mathbf{k}}) \text { is } \end{aligned} $$

If $A(0,3,4), B(1,5,6), C(-2,0,-2)$ are the vertices of a $\triangle A B C$ and the bisector of angle $A$ meets the side $B C$ at $D$, then $A D=$

If the direction cosines of two lines satisfy the equation $2 l+m-n=0, l^2-2 m^2+n^2=0$ and $\theta$ is the angle between the lines, then $\cos \theta=$

If the equation of the plane passing through the points $(2,1,2),(1,2,1)$ and perpendicular to the plane $2 x-y+2 z=1$ is $a x+b y+c z+d=0$, then $\frac{a+b}{c+d}=$