If the lines given by $$\frac{x-1}{2 \lambda}=\frac{y-1}{-5}=\frac{z-1}{2}$$ and $$\frac{x+2}{\lambda}=\frac{y+3}{\lambda}=\frac{z+5}{1}$$ are parallel, then the value of $$\lambda$$ is

The vector equation of the plane $\mathbf{r}=(2 \hat{\mathbf{i}}+\hat{\mathbf{k}})+\lambda(\hat{\mathbf{i}})+\mu(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-3 \hat{\mathbf{k}})$ in scalar product form is $\mathbf{r} \cdot(3 \hat{\mathbf{i}}+2 \hat{\mathbf{k}})=\alpha$, then $\alpha=\ldots$

The direction ratios of the normal to the plane passing through origin and the line of intersection of the planes $x+2 y+3 z=4$ and $4 x+3 y+2 z=1$ are $\ldots \ldots$

If line $\frac{2 x-4}{\lambda}=\frac{y-1}{2}=\frac{z-3}{1}$ and $\frac{x-1}{1}=\frac{3 y-1}{\lambda}=\frac{z-2}{1}$ are perpendicular to each other then $\lambda=$ ............