$A$ plane meets the coordinate axes at $A, B, C$ such that the centroid of the triangle $ABC$ is $(1, 2, 4)$. Then,the equation of the plane is

The distance between the two parallel planes $2x + y + 2z = 8$ and $4x + 2y + 4z + 5 = 0$ is: (in $/2$)

The vector equation of the plane $r = (2 \hat{i} + \hat{k}) + \lambda(\hat{i}) + \mu(\hat{i} + 2 \hat{j} - 3 \hat{k})$ in scalar product form is $r \cdot (3 \hat{i} + 2 \hat{k}) = \alpha$,then $\alpha = \dots$

If the planes $2x + 3y + 4z + 7 = 0$ and $4x + ky + 8z + 1 = 0$ are parallel,then the equation of the plane passing through the point $(k, k, k)$ and having the direction ratios of its normal as $(k-1, k, k+1)$ is

If the distance between the planes $2x + y + z + 1 = 0$ and $2x + y + z + \alpha = 0$ is $3$ units,then the product of all possible values of $\alpha$ is

If the plane $7x + 11y + 13z = 3003$ meets the coordinate axes at $A, B, C$,then the centroid of the $\triangle ABC$ is