In the following case,find the distance of the given point from the corresponding given plane.

Point	Plane
$(-6, 0, 0)$	$2x - 3y + 6z - 2 = 0$

Let the plane $x+3y-2z+6=0$ meet the coordinate axes at the points $A, B, C$. If the orthocentre of the triangle $ABC$ is $\left(\alpha, \beta, \frac{6}{7}\right)$,then $98(\alpha+\beta)^2$ is equal to $........$.

If the plane $\frac{x}{2} + \frac{y}{3} + \frac{z}{3} = 1$ intersects the coordinate axes at $A, B, C$,then the area of $\Delta ABC$ is:

The $x$-intercept of a plane $\pi$ passing through the point $(1,1,1)$ is $\frac{5}{2}$ and the perpendicular distance from the origin to the plane $\pi$ is $\frac{5}{7}$. If the $y$-intercept of the plane $\pi$ is negative and the $z$-intercept is positive,then its $y$-intercept is

Direction ratios of the normal to a plane passing through $(1, 0, 0)$ and $(0, 1, 0)$ which makes an angle of $\frac{\pi}{4}$ with the plane $x + y - 3 = 0$ are:

If the point $(2, \alpha, \beta)$ lies on the plane which passes through the points $(3, 4, 2)$ and $(7, 0, 6)$ and is perpendicular to the plane $2x - 5y = 15$,then $2\alpha - 3\beta$ is equal to: