If $\bar{p}=2 \hat{i}+\hat{k}, \bar{q}=\hat{i}+\hat{j}+\hat{k}, \bar{r}=4 \hat{i}-3 \hat{j}+7 \hat{k}$ and a vector $\overline{\mathrm{m}}$ is such that $\overline{\mathrm{m}} \times \overline{\mathrm{q}}=\overline{\mathrm{r}} \times \overline{\mathrm{q}}, \overline{\mathrm{m}} \cdot \overline{\mathrm{p}}=0$, then $\overline{\mathrm{m}}=\ldots$.

If the point $(1, \alpha, \beta)$ lies on the line of the shortest distance between the lines $\frac{x+2}{-3}=\frac{y-2}{4}=\frac{z-5}{2}$ and $\frac{x+2}{-1}=\frac{y+6}{2}, \mathrm{z}=1$, then $\alpha+\beta=$

The angle between the lines $x-3 y-4=0,4 y-z+5=0$ and $x+3 y-11=0,2 y-z+6=0$ is

If the area of parallelogram, whose diagonals are $\hat{\mathrm{i}}-\hat{\mathrm{j}}+2 \hat{\mathrm{k}}$ and $2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+\alpha \hat{\mathrm{k}}$ is $\frac{\sqrt{93}}{2}$ sq. units, then $\alpha=$