The coordinates of the foot of the perpendicular drawn from the origin to a plane is $(2, 4, -3)$. The equation of the plane is

If for some $\alpha$ and $\beta$ in $\mathbb{R},$ the intersection of the following three planes $x+4y-2z=1$,$x+7y-5z=\beta$,and $x+5y+\alpha z=5$ is a line in $\mathbb{R}^{3},$ then $\alpha+\beta$ is equal to

$A$ plane $\pi$ passing through the point $(1,1,1)$ is perpendicular to the line joining the points $(6,3,2)$ and $(1,-4,-9)$. If $ax+by+cz-23=0$ is the equation of the plane $\pi$,then $a+b-c=$

Find the equation of a plane that makes equal intercepts of unit length on the axes.

Let $\pi$ be the plane that passes through the point $(-2, 1, -1)$ and is parallel to the plane $2x - y + 2z = 0$. Then the foot of the perpendicular drawn from the point $(1, 2, 1)$ to the plane $\pi$ is

$A$ vector $\overrightarrow{V}$ in the first octant is inclined to the $x$-axis at $60^{\circ}$,to the $y$-axis at $45^{\circ}$ and to the $z$-axis at an acute angle. If a plane passing through the points $(\sqrt{2}, -1, 1)$ and $(a, b, c)$ is normal to $\overrightarrow{V}$,then: