The equation of the plane passing through the intersection of the planes $x + y + z = 1$ and $2x + 3y - z + 4 = 0$ and parallel to the $x$-axis is:

If the straight lines $\frac{x-1}{2}=\frac{y+1}{k}=\frac{z}{2}$ and $\frac{x+1}{5}=\frac{y+1}{2}=\frac{z}{k}$ are coplanar,then the plane$(s)$ containing these two lines is(are):

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:

What is the distance between the line $\frac{x - 1}{3} = \frac{y + 2}{-2} = \frac{z - 1}{2}$ and the plane $2x + 2y - z = 6$?

Find the distance of the point $(-1,-5,-10)$ from the point of intersection of the line $\vec{r}=2 \hat{i}-\hat{j}+2 \hat{k}+\lambda(3 \hat{i}+4 \hat{j}+2 \hat{k})$ and the plane $\vec{r} \cdot(\hat{i}-\hat{j}+\hat{k})=5$.

Show that the planes $\vec{r} \cdot (\hat{i} + 2\hat{j} + 2\hat{k}) = 19$ and $\vec{r} \cdot (4\hat{i} - 3\hat{j} + 12\hat{k}) + 3 = 0$ are perpendicular. Find the equation of the plane containing these two lines (Note: The question asks for the plane containing the intersection of these two planes).