The point of intersection of the line $\frac{x}{1} = \frac{y - 1}{2} = \frac{z + 2}{3}$ and the plane $2x + 3y + z = 0$ 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:

The line $\frac{x - 2}{3} = \frac{y - 3}{4} = \frac{z - 4}{5}$ is parallel to the plane:

The angle between the line $\vec{r}=(\hat{i}+\hat{j}-\hat{k})+\lambda(3\hat{i}+\hat{j})$ and the plane $\vec{r} \cdot (\hat{i}+2\hat{j}+3\hat{k})=8$ is:

If the equation of the plane passing through the line of intersection of the planes $2x - y + z = 3$ and $4x - 3y + 5z + 9 = 0$ and parallel to the line $\frac{x + 1}{-2} = \frac{y + 3}{4} = \frac{z - 2}{5}$ is $ax + by + cz + 6 = 0$,then $a + b + c$ is equal to $.............$.

Let $Q$ and $R$ be the feet of perpendiculars from the point $P(a, a, a)$ on the lines $x=y, z=1$ and $x=-y, z=-1$ respectively. If $\angle QPR$ is a right angle,then $12a^2$ is equal to