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

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 angle between the line $\frac{x + 1}{3} = \frac{y - 1}{4} = \frac{z - 2}{2}$ and the plane $2x - 3y + z + 4 = 0$ is:

The value of $m$ such that the line $\frac{x-4}{1}=\frac{y-2}{1}=\frac{z+m}{2}$ lies in the plane $2x-4y+z=7$ is

Let $A=(2,0,-1)$,$B=(1,-2,0)$,$C=(1,2,-1)$,and $D=(0,-1,-2)$ be four points. If $\theta$ is the acute angle between the plane determined by $A, B, C$ and the plane determined by $A, C, D$,then $\tan \theta=$

At what point does the line joining the points $(2, -3, 1)$ and $(3, -4, -5)$ intersect the plane $2x + y + z = 7$?