The foot of the perpendicular drawn from the origin to the plane $x+y+3z-4=0$ is

The equation of the plane passing through $3 \hat{i}+2 \hat{j}+6 \hat{k}$ and parallel to the vectors $2 \hat{i}+\hat{j}+\hat{k}$ and $\hat{i}-\hat{j}+\hat{k}$ is

Let $\alpha, \beta, \gamma, \delta$ be real numbers such that $\alpha^2+\beta^2+\gamma^2 \neq 0$ and $\alpha+\gamma=1$. Suppose the point $(3,2,-1)$ is the mirror image of the point $(1,0,-1)$ with respect to the plane $\alpha x+\beta y+\gamma z=\delta$. Then which of the following statements is/are $TRUE$?
$(A)$ $\alpha+\beta=2$
$(B)$ $\delta-\gamma=3$
$(C)$ $\delta+\beta=4$
$(D)$ $\alpha+\beta+\gamma=\delta$

If $(2,3,-1)$ is the foot of the perpendicular from $(4,2,1)$ to a plane,then the equation of the plane is

The equation of the plane in Cartesian form, which is at a distance of $\frac{6}{\sqrt{29}}$ from the origin and its normal vector drawn from the origin being $2 \hat{i}-3 \hat{j}+4 \hat{k}$, is

$A$ plane meets the coordinate axes at points $A$, $B$ and $C$, such that the centroid of triangle $ABC$ is $(2, -\frac{2}{3}, \frac{1}{2})$. The perpendicular distance from the origin to this plane is: