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

Find the equation of the plane passing through $(a, b, c)$ and parallel to the plane $\vec{r} \cdot(\hat{i}+\hat{j}+\hat{k})=2$.

The Cartesian equation of a plane which passes through the points $A(2,2,2)$ and makes equal non-zero intercepts on the coordinate axes is

If the angle between the planes $\bar{r} \cdot(11 \hat{i}-2 \hat{j}+\alpha \hat{k})=7$ and $\bar{r} \cdot(2 \hat{i}+4 \hat{j}-2 \hat{k})=5$ is $\frac{\pi}{2}$, then $\alpha=$

If $\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c}$ are three non-coplanar vectors, then the vector equation $\overrightarrow{r}=(1-p-q) \overrightarrow{a}+p \overrightarrow{b}+q \overrightarrow{c}$ represents a :

The distance between the planes given by $r \cdot (i + 2j - 2k) + 5 = 0$ and $r \cdot (i + 2j - 2k) - 8 = 0$ is