Direction ratios of the normal to a plane passing through $(1, 0, 0)$ and $(0, 1, 0)$ which makes an angle of $\frac{\pi}{4}$ with the plane $x + y - 3 = 0$ are:

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

Consider three planes:
$P_1: x-y+z=1$
$P_2: x+y-z=-1$
$P_3: x-3y+3z=2$
Let $L_1, L_2, L_3$ be the lines of intersection of the planes $P_2$ and $P_3$,$P_3$ and $P_1$,and $P_1$ and $P_2$,respectively.
$STATEMENT-1$: At least two of the lines $L_1, L_2$ and $L_3$ are non-parallel.
$STATEMENT-2$: The three planes do not have a common point.

The distance between two parallel planes $ax+by+cz+d_1=0$ and $ax+by+cz+d_2=0$ is given by $\frac{|d_1-d_2|}{\sqrt{a^2+b^2+c^2}}$. If the plane $2x-y+2z+3=0$ is at distances of $\frac{1}{3}$ and $\frac{2}{3}$ units from the planes $4x-2y+4z+\lambda=0$ and $2x-y+2z+\mu=0$ respectively,then the maximum value of $\lambda+\mu$ is:

The foot of the perpendicular drawn from the origin to the plane is $(4, -2, -5)$. Hence,the equation of the plane is

$A$ variable plane is at a constant distance $p$ from the origin and meets the axes in $A, B$ and $C$. The locus of the centroid of the tetrahedron $OABC$ is