The vector equation of the plane passing through the point $(2, 1, -1)$ and the line of intersection of the planes $r \cdot (i + 3j - k) = 0$ and $r \cdot (j + 2k) = 0$ is:

If the equation of the plane passing through the mirror image of a point $(2,3,1)$ with respect to the line $\frac{x+1}{2}=\frac{y-3}{1}=\frac{z+2}{-1}$ and containing the line $\frac{x-2}{3}=\frac{1-y}{2}=\frac{z+1}{1}$ is $\alpha x+\beta y+\gamma z=24$,then $\alpha+\beta+\gamma$ is equal to ..... .

The equation of the plane passing through the line $\frac{x - 1}{5} = \frac{y + 2}{6} = \frac{z - 3}{4}$ and the point $(4, 3, 7)$ is

The equation of the plane passing through the lines $\frac{x - 4}{1} = \frac{y - 3}{1} = \frac{z - 2}{2}$ and $\frac{x - 3}{1} = \frac{y - 2}{-4} = \frac{z}{5}$ is

The acute angle $\theta$ between the line $\vec{r} = (\hat{i} + 2\hat{j} + \hat{k}) + \lambda(\hat{i} + \hat{j} + \hat{k})$ and the plane $\vec{r} \cdot (2\hat{i} + p\hat{j} + \hat{k}) = 8$ is given by $\sin \theta = \frac{|\vec{b} \cdot \vec{n}|}{|\vec{b}| |\vec{n}|}$, where $\vec{b} = \hat{i} + \hat{j} + \hat{k}$ and $\vec{n} = 2\hat{i} + p\hat{j} + \hat{k}$. If $\theta = \sin^{-1} \left( \frac{\sqrt{2}}{3} \right)$, then the value$(s)$ of $p$ is/are:

The line $\frac{x - 3}{2} = \frac{y - 4}{3} = \frac{z - 5}{4}$ lies in the plane $4x + 4y - kz - d = 0$. The values of $k$ and $d$ are