Let two vertices of triangle $ABC$ be $(2,4,6)$ and $(0,-2,-5)$,and its centroid be $(2,1,-1)$. If the image of the third vertex in the plane $x+2y+4z=11$ is $(\alpha, \beta, \gamma)$,then $\alpha \beta+\beta \gamma+\gamma \alpha$ is equal to

The line passing through the points $(1, 1, -1)$ and $(3, -1, 0)$ makes an angle of $\operatorname{Tan}^{-1}\left(\frac{1}{\sqrt{8}}\right)$ with the plane $\sqrt{\lambda} x + 3y + 6z = 17$. Then $\lambda =$

Let the plane $P: \vec{r} \cdot \vec{a} = d$ contain the line of intersection of two planes $\vec{r} \cdot (\hat{i} + 3\hat{j} - \hat{k}) = 6$ and $\vec{r} \cdot (-6\hat{i} + 5\hat{j} - \hat{k}) = 7$. If the plane $P$ passes through the point $(2, 3, 1/2)$,then the value of $\frac{|13\vec{a}|^2}{d^2}$ is equal to

Let $Q$ be the mirror image of the point $P(1, 2, 1)$ with respect to the plane $x + 2y + 2z = 16$. Let $T$ be a plane passing through the point $Q$ and containing the line $\vec{r} = -\hat{k} + \lambda(\hat{i} + \hat{j} + 2\hat{k}), \lambda \in R$. Then,which of the following points lies on $T$?

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

The lines $\frac{x-2}{1}=\frac{y-3}{1}=\frac{z-4}{-k}$ and $\frac{x-1}{k}=\frac{y-4}{2}=\frac{z-5}{1}$ are coplanar if