Equation of the plane containing the line $x + 2y + 3z - 5 = 0 = 3x + 2y + z - 5$ which is parallel to the line $x - 1 = 2 - y = z - 3$ is:

Let $P$ be a plane containing the line $\frac{x-1}{3}=\frac{y+6}{4}=\frac{z+5}{2}$ and parallel to the line $\frac{x-3}{4}=\frac{y-2}{-3}=\frac{z+5}{7}$. If the point $(1, -1, \alpha)$ lies on the plane $P$,then the value of $|5\alpha|$ is equal to ....... .

If the equation of the plane passing through the line of intersection of the planes $2x - 7y + 4z - 3 = 0$ and $3x - 5y + 4z + 11 = 0$ and the point $(-2, 1, 3)$ is $ax + by + cz - 7 = 0$,then the value of $2a + b + c - 7$ is

Find the vector equation of the line passing through $(1, 2, 3)$ and parallel to the planes $\vec{r} \cdot (\hat{i} - \hat{j} + 2\hat{k}) = 5$ and $\vec{r} \cdot (3\hat{i} + \hat{j} + \hat{k}) = 6$.

Find the angle between the line $\vec{r} = (2\hat{i} - \hat{j} + \hat{k}) + \lambda(-\hat{i} + \hat{j} + \hat{k})$ and the plane $\vec{r} \cdot (3\hat{i} + 2\hat{j} - \hat{k}) = 4$.

If the lines $\frac{x - 2}{1} = \frac{y - 3}{1} = \frac{z - 4}{-k}$ and $\frac{x - 1}{k} = \frac{y - 4}{2} = \frac{z - 5}{1}$ are coplanar,then $k$ can have: