The equation of the plane passing through the intersection of the planes $x + y + z = 6$ and $2x + 3y + 4z + 5 = 0$ and the point $(1, 1, 1)$ is:

Let the equation of the plane passing through the line $x-2y-z-5=0=x+y+3z-5$ and parallel to the line $x+y+2z-7=0=2x+3y+z-2$ be $ax+by+cz=65$. Then the distance of the point $(a, b, c)$ from the plane $2x+2y-z+16=0$ is $..........$.

In $R^3$,consider the planes $P_1: y=0$ and $P_2: x+z=1$. Let $P_3$ be a plane,different from $P_1$ and $P_2$,which passes through the intersection of $P_1$ and $P_2$. If the distance of the point $(0,1,0)$ from $P_3$ is $1$ and the distance of a point $(\alpha, \beta, \gamma)$ from $P_3$ is $2$,then which of the following relations is (are) true?
$(A)$ $2\alpha+\beta+2\gamma+2=0$
$(B)$ $2\alpha-\beta+2\gamma+4=0$
$(C)$ $2\alpha+\beta-2\gamma-10=0$
$(D)$ $2\alpha-\beta+2\gamma-8=0$

Find the ratio in which the plane $x - 2y + 3z = 17$ divides the line segment joining the points $(-2, 4, 7)$ and $(3, -5, 8)$.

The plane $3x + 4y + 6z + 7 = 0$ is rotated about the line $r = (\hat{i} + 2\hat{j} - 3\hat{k}) + t(2\hat{i} - 3\hat{j} + \hat{k})$ until the plane passes through the origin. The equation of the plane in the new position is

The lines $\frac{x-0}{1}=\frac{y-2}{2}=\frac{z+3}{\lambda}$ and $\frac{x-2}{2}=\frac{y-6}{3}=\frac{z-3}{\lambda}$ are coplanar. If $p$ is the plane containing these lines,then which of the following points lies on the plane for all values of $\lambda$?