Find the vector equation of the plane passing through the points $\hat{i} + \hat{j} - 2\hat{k}$,$2\hat{i} - \hat{j} + \hat{k}$,and $\hat{i} + 2\hat{j} + \hat{k}$.

The equation of the plane passing through $(2, 3, 4)$ and parallel to the plane $5x - 6y + 7z = 3$ is:

$A$ plane meets the coordinate axes at $A, B, C$ such that the centroid of the $\Delta ABC$ is the point $(\alpha, \beta, \gamma)$. Show that the equation of the plane is $\frac{x}{\alpha} + \frac{y}{\beta} + \frac{z}{\gamma} = 3$.

$A$ plane $\pi_1$ passing through the point $3 \hat{i}-7 \hat{j}+5 \hat{k}$ is perpendicular to the vector $\hat{i}+2 \hat{j}-2 \hat{k}$ and another plane $\pi_2$ passing through the point $2 \hat{i}+7 \hat{j}-8 \hat{k}$ is perpendicular to the vector $3 \hat{i}+2 \hat{j}+6 \hat{k}$. If $p_1$ and $p_2$ are the perpendicular distances from the origin to the planes $\pi_1$ and $\pi_2$ respectively, then $p_1-p_2=$

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 foot of the perpendicular drawn from $(0,0,0)$ to a plane is $(1,2,3)$,then the equation of the plane is: