Three Dimensional Geometry

If the image of the point $P(1, 0, 3)$ in the line joining the points $A(4, 7, 1)$ and $B(3, 5, 3)$ is $Q(\alpha, \beta, \gamma)$, then $\alpha + \beta + \gamma$ is equal to:

The line $L_1$ is parallel to the vector $\vec{a} = -3\hat{i} + 2\hat{j} + 4\hat{k}$ and passes through the point $(7, 6, 2)$ and the line $L_2$ is parallel to the vector $\vec{b} = 2\hat{i} + \hat{j} + 3\hat{k}$ and passes through the point $(5, 3, 4)$. The shortest distance between the lines $L_1$ and $L_2$ is:

Let the vertices Q and R of the triangle PQR lie on the line $\frac{x+3}{5}=\frac{y-1}{2}=\frac{z+4}{3}$, QR = 5, and the coordinates of the point P be $(0,2,3)$. If the area of the triangle PQR is $\tfrac{m}{n}$, then:

Each of the angles $\beta$ and $\gamma$ that a given line makes with the positive $y$- and $z$-axes, respectively, is half of the angle that this line makes with the positive $x$-axis. Then the sum of all possible values of the angle $\beta$ is:

The distance of the point $(7,10,11)$ from the line $\displaystyle\frac{x-4}{1}=\frac{y-4}{0}=\frac{z-2}{3}$ along the line $\displaystyle\frac{x-9}{2}=\frac{y-13}{-3}=\frac{z-17}{6}$ is:

Line L1 passes through the point (1, 2, 3) and is parallel to the z-axis. Line L2 passes through the point (λ, 5, 6) and is parallel to the y-axis. If for λ = λ₁, λ₂ with λ₂ < λ₁ the shortest distance between these two lines is 3, then the square of the distance of the point (λ₁, λ₂, 7) from the line L1 is:

Let the values of $p$, for which the shortest distance between the lines $\displaystyle\frac{x+1}{3}=\frac{y}{4}=\frac{z}{5}$ and $\mathbf{r}=(p\hat{i}+2\hat{j}+\hat{k})+\lambda(2\hat{i}+3\hat{j}+4\hat{k})$ is $\tfrac{1}{\sqrt6}$, be $a,b$ with $a<b$. Then the length of the latus rectum of the ellipse $\displaystyle\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is:

Let $A$ be the point of intersection of the lines $L_1:\;\frac{x-7}{1}=\frac{y-5}{0}=\frac{z-3}{-1},$ and $L_2:\;\frac{x-1}{3}=\frac{y+3}{4}=\frac{z+7}{5}.$Let $B$ and $C$ be points on $L_1$ and $L_2$ respectively such that $AB=AC=\sqrt{15}$. Then the square of the area of triangle $ABC$ is:

Let the shortest distance between the lines $\frac{x-3}{3} = \frac{y-\alpha}{-1} = \frac{z-3}{1}$ and $\frac{x+3}{-3} = \frac{y+7}{2} = \frac{z-\beta}{4}$ be 3\sqrt{30}. Then the positive value of 5\alpha + \beta is:

Let A and B be two distinct points on the line L: (x-6)/3 = (y-7)/2 = (z-7)/-2. Both A and B are at a distance 2\sqrt{17} from the foot of the perpendicular drawn from the point (1, 2, 3) to the line L. If O is the origin, then OA \cdot OB is equal to: