The angle between the line $$r =(\hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}})+\lambda(3 \hat{\mathbf{i}}+\hat{\mathbf{j}})$$ and the plane $$\mathbf{r} \cdot(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}})=8$$ is

The direction cosines of a line which is perpendicular to lines whose direction ratios are $$3,-2,4$$ and $$1,3,-2$$ are

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 $\alpha=\ldots$