The equation of a plane passing through the line of intersection of the planes $x+2y+3z=2$ and $x-y+z=3$ and at a distance $\frac{2}{\sqrt{3}}$ from the point $(3,1,-1)$ is

Let the image of the point $P(1, 6, a)$ in the line $L: \frac{x-1}{2} = \frac{y-2}{b} = \frac{z-a+1}{1}, b>0$, be $Q(\frac{a}{3}, 0, a+c)$. If $S(\alpha, \beta, \gamma), \alpha > 0$, is the point on $L$ such that the distance of $S$ from the foot of perpendicular $F$ from the point $P$ on $L$ is $2\sqrt{14}$, then $\alpha + \beta + \gamma$ is equal to:

If the line $r = a + t b$ is parallel to the plane $r = c + l d + m e$, then

If the lines $x = 1 + s, y = -3 - \lambda s, z = 1 + \lambda s$ and $x = t/2, y = 1 + t, z = 2 - t$ are coplanar,find the value of $\lambda$.

If the straight lines $x = 1 + s, y = -3 - \lambda s, z = 1 + \lambda s$ and $x = t/2, y = 1 + t, z = 2 - t$,with parameters $s$ and $t$ respectively,are coplanar,then $\lambda$ equals

The coordinates of the point where the line joining $(1, 1, 1)$ and $(2, 2, 2)$ intersects the plane $x + y + z = 9$ are: