The equation of a plane containing the lines $\overline{r}=(\hat{\imath}+2 \hat{\jmath}-4 \hat{k})+\lambda(2 \hat{\imath}+3 \hat{\jmath}+6 \hat{k})$ and $\overline{r}=(\hat{\imath}+3 \hat{\jmath}+4 \hat{k})+\mu(\hat{\imath}+\hat{\jmath}-\hat{k})$ is

$l, m, n$ are three unit vectors in a right-handed system and $L$ is a line through the points $A, B, C$ whose position vectors are $p l + 7 m - 6 n, 2 l + 5 m - 4 n$ and $l + 4 m - 3 n$ respectively. If the equation of the plane containing $L$ and the point $(-p, p, p+1)$ is $ax + by + cz = 1$, then $p(a+b+c) =$

Let a plane $P$ pass through the point $A(3, 7, -7)$ and contain the line $L: \frac{x - 2}{-3} = \frac{y - 3}{2} = \frac{z + 2}{1}$. If the distance of the plane $P$ from the origin is $d$, then $d^2$ 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 $...........$.

Let $P_1: 2x + y - z = 3$ and $P_2: x + 2y + z = 2$ be two planes. Then,which of the following statement$(s)$ is (are) $TRUE$?
$(A)$ The line of intersection of $P_1$ and $P_2$ has direction ratios $1, -1, 1$.
$(B)$ The line $\frac{3x - 4}{9} = \frac{1 - 3y}{9} = \frac{z}{3}$ is perpendicular to the line of intersection of $P_1$ and $P_2$.
$(C)$ The acute angle between $P_1$ and $P_2$ is $60^{\circ}$.
$(D)$ If $P_3$ is the plane passing through the point $(4, 2, -2)$ and perpendicular to the line of intersection of $P_1$ and $P_2$,then the distance of the point $(2, 1, 1)$ from the plane $P_3$ is $\frac{2}{\sqrt{3}}$.

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