What is the reflection of the point $P(1, 3, 4)$ in the plane $2x - y + z + 3 = 0$?

The ratio in which the line joining $(2, -4, 3)$ and $(-4, 5, -6)$ is divided by the plane $3x + 2y + z - 4 = 0$ is

If $\vec{a}=\hat{i}-\hat{j}+3 \hat{k}$ and $\vec{c}=-\hat{k}$ are position vectors of two points, and $\vec{b}=2 \hat{i}-\hat{j}+\lambda \hat{k}$ and $\vec{d}=\hat{i}+2 \hat{j}-\hat{k}$ are two vectors, then the lines $\vec{r}=\vec{a}+t \vec{b}$ and $\vec{r}=\vec{c}+s \vec{d}$ are:

The distance between the line $\vec{r} = (2\hat{i} - 2\hat{j} + 3\hat{k}) + \lambda(\hat{i} - \hat{j} + 4\hat{k})$ and the plane $\vec{r} \cdot (\hat{i} + 5\hat{j} + \hat{k}) = 5$ is:

For real numbers $\alpha$ and $\beta \neq 0$,if the point of intersection of the straight lines $\frac{x-\alpha}{1}=\frac{y-1}{2}=\frac{z-1}{3}$ and $\frac{x-4}{\beta}=\frac{y-6}{3}=\frac{z-7}{3}$ lies on the plane $x+2y-z=8$,then $\alpha-\beta$ is equal to:

If the lines $\frac{x-1}{2}=\frac{y+1}{k}=\frac{z}{2}$ and $\frac{x+1}{5}=\frac{y+1}{2}=\frac{z}{k}$ are coplanar,then the equation of the plane containing these lines is: