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 $\frac{x+2}{-1}=\frac{y+6}{2}, z=1$,then $\alpha+\beta=$

The ratio in which the plane $x - 2y + 3z = 17$ divides the line segment joining the points $A(-2, 4, 7)$ and $B(3, -5, 8)$ is:

Let $P(x_1, y_1, z_1)$ be the foot of the perpendicular drawn from the point $Q(2, -2, 1)$ to the plane $x - 2y + z = 1$. If $d$ is the perpendicular distance from the point $Q$ to the plane and $l = x_1 + y_1 + z_1$,then the value of $l + 3d^2$ is:

The coordinates of the point where the line $\frac{x-1}{2}=\frac{y-2}{-3}=\frac{z+3}{4}$ meets the plane $2x+4y-z=1$ are

Three lines are given by $\overrightarrow{r} = \lambda \hat{i}, \lambda \in R$,$\overrightarrow{r} = \mu(\hat{i} + \hat{j}), \mu \in R$ and $\overrightarrow{r} = v(\hat{i} + \hat{j} + \hat{k}), v \in R$. Let the lines cut the plane $x + y + z = 1$ at the points $A, B$ and $C$ respectively. If the area of the triangle $ABC$ is $\Delta$,then the value of $(6 \Delta)^2$ equals.

The lines $x = ay - 1 = z - 2$ and $x = 3y - 2 = bz - 2$ $(ab \neq 0)$ are coplanar,if: