Find the equation of the plane containing the line $\frac{x + 1}{-3} = \frac{y - 3}{2} = \frac{z + 2}{1}$ and the point $(0, 7, -7)$.

Show that the lines $\frac{x-a+d}{\alpha-\delta}=\frac{y-a}{\alpha}=\frac{z-a-d}{\alpha+\delta}$ and $\frac{x-b+c}{\beta-\gamma}=\frac{y-b}{\beta}=\frac{z-b-c}{\beta+\gamma}$ are coplanar.

If the line $\frac{2x - 8}{\sin \beta} = \frac{y - \sin \alpha}{1} = \frac{z - 1}{\cos \alpha}$,where $\beta \in R$ and $\sin \beta \neq 1$,lies in the plane $2x - (\sin \beta)y + (\cos \beta)z = k$ for all $\alpha \in R$,then:

The distance of the point $(7,5,2)$ from the plane $3x+4y+z-8=0$ measured parallel to the line $\frac{x-1}{3}=\frac{y-2}{6}=\frac{z+1}{2}$ is:

The equation of the line passing through the point $(2, 3, 1)$ and parallel to the line of intersection of the planes $x - 2y - z + 5 = 0$ and $x + y + 3z = 6$ is:

The number of distinct real values of $\lambda$ for which the lines $\frac{x - 1}{1} = \frac{y - 2}{2} = \frac{z + 3}{\lambda^2}$ and $\frac{x - 3}{1} = \frac{y - 2}{\lambda^2} = \frac{z - 1}{2}$ are coplanar is