If the line $\vec{r} = (\hat{i} - 2\hat{j} - \hat{k}) + \lambda (2\hat{i} + \hat{j} + 2\hat{k})$ is parallel to the plane $\vec{r} \cdot (3\hat{i} - 2\hat{j} - m\hat{k}) = 14$,then the value of $m$ is:

The position vectors of points $A$ and $B$ are $i - j + 3k$ and $3i + 3j + 3k$ respectively. The equation of a plane is $r \cdot (5i + 2j - 7k) + 9 = 0$. The points $A$ and $B$:

If the line $\frac{2x - 8}{\sin \beta} = \frac{y - \sin \alpha}{1} = \frac{z - 1}{\cos \alpha}$,where $\beta \in R$ and $\sin \beta \neq 1$,lies in the plane $2x - (\sin \beta)y + (\cos \beta)z = k$ for all $\alpha \in R$,then:

The line $\frac{x - 2}{3} = \frac{y - 3}{4} = \frac{z - 4}{5}$ is parallel to the plane:

Find the image of the point having position vector $\hat{i}+3 \hat{j}+4 \hat{k}$ in the plane $\vec{r} \cdot(2 \hat{i}-\hat{j}+\hat{k})+3=0$.

Find the equation of the plane passing through the line of intersection of the planes $\vec{r} \cdot(\hat{i}+\hat{j}+\hat{k})=1$ and $\vec{r} \cdot(2 \hat{i}+3 \hat{j}-\hat{k})+4=0$ and parallel to the $x$-axis.