In three-dimensional space,the equation $3y + 4z = 0$ represents:

If the plane $\frac{x}{2} + \frac{y}{3} + \frac{z}{3} = 1$ intersects the coordinate axes at $A, B, C$,then the area of $\Delta ABC$ is:

The perpendicular distance from the origin to the plane containing the points having position vectors $\hat{i}+2\hat{j}+3\hat{k}$,$2\hat{i}+3\hat{j}-4\hat{k}$,and $3\hat{i}-4\hat{j}+5\hat{k}$ is

$A$ variable plane $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$,which is at a unit distance from the origin,cuts the coordinate axes at $A, B$,and $C$. If the centroid $(x, y, z)$ of $\triangle ABC$ satisfies $\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=k$,then $k$ equals:

If a plane cuts off intercepts $-6, 3, 4$ from the coordinate axes,then the length of the perpendicular from the origin to the plane is

If the points $(1, -1, \lambda)$ and $(-3, 0, 1)$ are equidistant from the plane $3x - 4y - 12z + 13 = 0$,then the sum of all possible values of $\lambda$ is