What is the reflection of the point $P(1, 3, 4)$ in the plane $2x - y + z + 3 = 0$?

Let the line $\ell: x = \frac{1-y}{-2} = \frac{z-3}{\lambda}, \lambda \in R$ meet the plane $P: x + 2y + 3z = 4$ at the point $(\alpha, \beta, \gamma)$. If the angle between the line $\ell$ and the plane $P$ is $\cos^{-1}\left(\sqrt{\frac{5}{14}}\right)$,then $\alpha + 2\beta + 6\gamma$ is equal to

If the line $\bar{r}=(\hat{i}-2 \hat{j}+3 \hat{k})+\lambda(2 \hat{i}+\hat{j}+2 \hat{k})$ is parallel to the plane $\bar{r} \cdot(3 \hat{i}-2 \hat{j}-m \hat{k})=5$,then the value of $m$ is:

The acute angle between the line $\bar{r}=(\hat{i}+2\hat{j}+\hat{k})+\lambda(\hat{i}+\hat{j}+\hat{k})$ and the plane $\bar{r} \cdot(2\hat{i}-\hat{j}+\hat{k})=5$ is

The plane $2x - y + 3z + 5 = 0$ is rotated through $90^o$ about its line of intersection with the plane $5x - 4y - 2z + 1 = 0$. The equation of the plane in its new position is:

Let $P$ be a plane passing through the points $(1,0,1), (1,-2,1)$ and $(0,1,-2)$. Let a vector $\vec{a} = \alpha \hat{i} + \beta \hat{j} + \gamma \hat{k}$ be such that $\vec{a}$ is parallel to the plane $P$,perpendicular to $(\hat{i} + 2 \hat{j} + 3 \hat{k})$ and $\vec{a} \cdot (\hat{i} + \hat{j} + 2 \hat{k}) = 2$. Then $(\alpha - \beta + \gamma)^2$ equals: