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:

The equation of the straight line passing through $(1, 2, 3)$ and perpendicular to the plane $x + 2y - 5z + 9 = 0$ is

The position vector of the point in which the line joining the points $i - 2j + k$ and $3k - 2j$ cuts the plane through the origin and the points $4j$ and $2i + k$,is

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:

Find the equation of the plane passing through the points $(2, 1, -1)$ and $(-1, 3, 4)$ and perpendicular to the plane $x - 2y + 4z = 10$.

Let $P_{1}: \vec{r} \cdot(2 \hat{i} + \hat{j} - 3 \hat{k}) = 4$ be a plane. Let $P_{2}$ be another plane which passes through the points $(2, -3, 2)$,$(2, -2, -3)$,and $(1, -4, 2)$. If the direction ratios of the line of intersection of $P_{1}$ and $P_{2}$ are $16, \alpha, \beta$,then the value of $\alpha + \beta$ is equal to