The distance between the line $\bar{r} = 3\hat{i} - 2\hat{j} + \hat{k} + \lambda(\hat{i} - 2\hat{j})$ and the plane $\bar{r} \cdot (2\hat{i} + \hat{j} + \hat{k}) = 4$ is

The point of intersection of the line $\frac{x}{1} = \frac{y - 1}{2} = \frac{z + 2}{3}$ and the plane $2x + 3y + z = 0$ is

The symmetric equation of the line formed by the intersection of the planes $3x + 2y + z - 5 = 0$ and $x + y - 2z - 3 = 0$ is:

Let the equation of the plane,that passes through the point $(1,4,-3)$ and contains the line of intersection of the planes $3x-2y+4z-7=0$ and $x+5y-2z+9=0$,be $\alpha x+\beta y+\gamma z+3=0$. Then $\alpha+\beta+\gamma$ is equal to:

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 $\frac{x-2}{3}=\frac{y+1}{-2}=\frac{z+3}{-1}$ lie on the plane $px-qy+z=5$,for some $p, q \in R$. The shortest distance of the plane from the origin is