The foot of the perpendicular drawn from $A(1, 2, 2)$ onto the plane $x+2y+2z-5=0$ is $B(\alpha, \beta, \gamma)$. If $\pi(x, y, z) \equiv x+2y+2z+5=0$ is a plane, then $-\pi(A) : \pi(B) =$ ?

Let the planes be $3x - 6y - 2z = 15$ and $2x + y - 2z = 5$.
Statement-$1$: The parametric equations of the line of intersection of the given planes are $x = 3 + 14t, y = 1 + 2t, z = 15t$.
Statement-$2$: The vector $14\hat{i} + 2\hat{j} + 15\hat{k}$ is parallel to the line of intersection of the given planes.

Let $\Pi$ be a plane containing the points $(0,-5,-1), (1,-2,5), (-3,5,0)$ and $L$ be a line passing through the point $(0,-5,-1)$ and parallel to the vector $\hat{i}+5\hat{j}-6\hat{k}$. Then the length of the projection of the unit normal vector to the plane $\Pi$ on the line $L$ is

The equation of the line,passing through $(1, 2, 3)$ and parallel to the planes $x - y + 2z = 5$ and $3x + y + z = 6$,is

Let $\theta$ be the angle between the planes $P_1=\vec{r} \cdot(\hat{i}+\hat{j}+2\hat{k})=9$ and $P_2=\vec{r} \cdot(2\hat{i}-\hat{j}+\hat{k})=15$. Let $L$ be the line that meets $P_2$ at the point $(4,-2,5)$ and makes an angle $\theta$ with the normal of $P_2$. If $\alpha$ is the angle between $L$ and $P_2$,then $(\tan^2 \theta)(\cot^2 \alpha)$ is equal to $...........$.

$\vec{r} \cdot(\hat{i}-\hat{j}+\hat{k})=5$ and $\vec{r} \cdot(2 \hat{i}+\hat{j}-\hat{k})=3$ are two planes. $A$ plane $\pi$ passing through the line of intersection of these two planes, passes through the point $(0,1,2)$. If the equation of $\pi$ is $\vec{r} \cdot(a \hat{i}+b \hat{j}+c \hat{k})=m$, then $\frac{b c}{a^2}=$