The equation of the plane containing the line $\frac{x}{1}=\frac{y}{2}=\frac{z}{3}$ and perpendicular to the plane containing the lines $\frac{x}{2}=\frac{y}{3}=\frac{z}{1}$ and $\frac{x}{3}=\frac{y}{2}=\frac{z}{1}$ is

$A$ line $L$ is passing through the point $A$ whose position vector is $\hat{i}+2 \hat{j}-3 \hat{k}$ and is parallel to the vector $2 \hat{i}+\hat{j}+2 \hat{k}$. $A$ plane $\pi$ is passing through the points $\hat{i}+\hat{j}+\hat{k}$ and $\hat{i}-\hat{j}-\hat{k}$ and is parallel to the vector $\hat{i}-2 \hat{j}$. Then the point where this plane $\pi$ meets the line $L$ is

The line of intersection of the planes $x + 2y = 0$ and $y - 3z + 3 = 0$ is

Let $P_{1}: \vec{r} \cdot(2 \hat{i} + \hat{j} - 3 \hat{k}) = 4$ be a plane. Let $P_{2}$ be another plane which passes through the points $(2, -3, 2)$,$(2, -2, -3)$,and $(1, -4, 2)$. If the direction ratios of the line of intersection of $P_{1}$ and $P_{2}$ are $16, \alpha, \beta$,then the value of $\alpha + \beta$ is equal to

Let $\pi_1$ be the plane determined by the vectors $\hat{i}+\hat{j}$ and $\hat{j}+\hat{k}$,and $\pi_2$ be the plane determined by the vectors $\hat{i}-\hat{j}$ and $\hat{i}+\hat{j}-\hat{k}$. Let $\vec{a}$ be a vector parallel to the line of intersection of $\pi_1$ and $\pi_2$. If $|\vec{a}|=\sqrt{14}$,then $|\vec{a} \cdot(\hat{i}+\hat{j}+\hat{k})|=$

If the line $x = y = z$ intersects the line defined by the equations $x \sin A + y \sin B + z \sin C - 18 = 0$ and $x \sin 2A + y \sin 2B + z \sin 2C - 9 = 0$,where $A, B, C$ are the angles of a triangle $ABC$,then $80 \left( \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} \right)$ is equal to $..........$.