What is the equation of the line passing through $(1, 2, 3)$ and perpendicular to the plane $3x + 4y - 5z = 6$?

The vector equation of the plane passing through the line of intersection of the planes $x + y + z = 1$ and $2x + 3y + 4z = 5$ which is perpendicular to the plane $x - y + z = 0$ is

Let the coordinates of one vertex of $\triangle ABC$ be $A(0, 2, \alpha)$ and the other two vertices lie on the line $\frac{x+\alpha}{5} = \frac{y-1}{2} = \frac{z+4}{3}$. For $\alpha \in \mathbb{Z}$,if the area of $\triangle ABC$ is $21$ sq. units and the line segment $BC$ has length $2\sqrt{21}$ units,then $\alpha^2$ is equal to $...........$.

$A$ ray of light passing through the point $A(1, 2, 3)$ strikes the plane $x+y+z=12$ at $B$ and on reflection passes through $C(3, 5, 9)$,then $OB$ is equal to

Let $P$ be the plane,which contains the line of intersection of the planes $x + y + z - 6 = 0$ and $2x + 3y + z + 5 = 0$ and it is perpendicular to the $xy$-plane. Then the distance of the point $(0, 0, 256)$ from $P$ is equal to

The line of intersection of the planes $x + 2y = 0$ and $y - 3z + 3 = 0$ is