The equation of the plane containing the line $\frac{x+1}{-3}=\frac{y-3}{2}=\frac{z+2}{1}$ and the point $(0,7,-7)$ is

The vector equation of a plane passing through the line of intersection of the planes $\overline{r} \cdot(\overline{i}-2 \overline{k})=3$ and $\overline{r} \cdot(2 \overline{j}+\overline{k})=5$,and passing through the point $\overline{i}+2 \overline{j}+3 \overline{k}$,is:

Let the coordinates of one vertex of $\triangle ABC$ be $A(0, 2, \alpha)$ and the other two vertices lie on the line $\frac{x+\alpha}{5} = \frac{y-1}{2} = \frac{z+4}{3}$. For $\alpha \in \mathbb{Z}$,if the area of $\triangle ABC$ is $21$ sq. units and the line segment $BC$ has length $2\sqrt{21}$ units,then $\alpha^2$ is equal to $...........$.

The distance of the point $(1, -2, 4)$ from the plane passing through the point $(1, 2, 2)$ and perpendicular to the planes $x - y + 2z = 3$ and $2x - 2y + z + 12 = 0$ is:

The angle between the line $r = (i + 2j - k) + \lambda (i - j + k)$ and the normal to the plane $r \cdot (2i - j + k) = 4$ is

Let $L$ be a line passing through the points $2 \hat{i}+3 \hat{j}+8 \hat{k}$ and $\hat{i}+6 \hat{j}+4 \hat{k}$. Let $P$ be a plane passing through $-5 \hat{i}+19 \hat{j}-14 \hat{k}$ and parallel to the vectors $\hat{i}-\hat{j}+\hat{k}$ and $\hat{i}-2 \hat{j}+3 \hat{k}$. If $L$ meets the plane $P$ at a point $A$,then the position vector of $A$ is: