If the line $\frac{x - x_1}{l} = \frac{y - y_1}{m} = \frac{z - z_1}{n}$ is parallel to the plane $ax + by + cz + d = 0$,then:

In what ratio does the plane $ax + by + cz + d = 0$ divide the line segment joining the points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$?

The line $\frac{x-2}{3}=\frac{y-1}{-5}=\frac{z+2}{2}$ lies in the plane $x+3y-\alpha z+\beta=0$. Then the value of $\alpha \beta$ is:

$A$ plane $P$ contains the line of intersection of the planes $\vec{r} \cdot (\hat{i}+\hat{j}+\hat{k}) = 6$ and $\vec{r} \cdot (2\hat{i}+3\hat{j}+4\hat{k}) = -5$. If $P$ passes through the point $(0, 2, -2)$,then the square of the distance of the point $(12, 12, 18)$ from the plane $P$ is

Let $A(2,5,7)$ be the image of the point $B(1,-2,3)$ with respect to a plane $\pi$. Let $C$ be the point where $AB$ meets the plane $\pi$. Let $D=(2,1,6)$. Then the direction cosines of $CD$ are

Let the plane containing the line of intersection of the planes $P_1: x+(\lambda+4)y+z=1$ and $P_2: 2x+y+z=2$ pass through the points $(0,1,0)$ and $(1,0,1)$. Then the distance of the point $(2\lambda, \lambda, -\lambda)$ from the plane $P_2$ is (in $\sqrt{6}$)