Let $\mathrm{P}$ and $\mathrm{Q}$ be the points on the line $\frac{x+3}{8}=\frac{y-4}{2}=\frac{z+1}{2}$ which are at a distance of 6 units from the point $\mathrm{R}(1,2,3)$. If the centroid of the triangle PQR is $(\alpha ,\beta ,\gamma )$, then ${\alpha }^2+{\beta }^2+{\gamma }^2$ is :

If the mirror image of the point $P(3,4,9)$ in the line $\frac{x-1}{3}=\frac{y+1}{2}=\frac{z-2}{1}$ is $(\alpha ,\beta ,\gamma )$, then 14 $(\alpha +\beta +\gamma )$ is :

If the shortest distance between the lines $\frac{x-\lambda }{-2}=\frac{y-2}{1}=\frac{z-1}{1}$ and $\frac{x-\sqrt{3}}{1}=\frac{y-1}{-2}=\frac{z-2}{1}$ is 1 , then the sum of all possible values of $\lambda$ is :

The distance, of the point $(7,-2,11)$ from the line $\frac{x-6}{1}=\frac{y-4}{0}=\frac{z-8}{3}$ along the line $\frac{x-5}{2}=\frac{y-1}{-3}=\frac{z-5}{6}$, is :

If the shortest distance between the lines $\frac{x-4}{1}=\frac{y+1}{2}=\frac{z}{-3}$ and $\frac{x-\lambda }{2}=\frac{y+1}{4}=\frac{z-2}{-5}$ is $\frac{6}{\sqrt{5}}$, then the sum of all possible values of $\lambda$ is :

Let the image of the point $$(1,0,7)$$ in the line $$\frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3}$$ be the point $$(\alpha ,\beta ,\gamma )$$. Then which one of the following points lies on the line passing through $$(\alpha ,\beta ,\gamma )$$ and making angles $$\frac{2\pi }{3}$$ and $$\frac{3\pi }{4}$$ with $$y$$-axis and $$z$$-axis respectively and an acute angle with $$x$$-axis ?

Let $$(\alpha ,\beta ,\gamma )$$ be the mirror image of the point $$(2,3,5)$$ in the line $$\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$$. Then, $$2\alpha +3\beta +4\gamma$$ is equal to

The shortest distance, between lines $$L_1$$ and $$L_2$$, where $$L_1:\frac{x-1}{2}=\frac{y+1}{-3}=\frac{z+4}{2}$$ and $$L_2$$ is the line, passing through the points $$\mathrm{A}(-4,4,3),\mathrm{B}(-1,6,3)$$ and perpendicular to the line $$\frac{x-3}{-2}=\frac{y}{3}=\frac{z-1}{1}$$, is

Let $$O$$ be the origin and the position vectors of $$A$$ and $$B$$ be $$2\hat{i}+2\hat{j}+\hat{k}$$ and $$2\hat{i}+4\hat{j}+4\hat{k}$$ respectively. If the internal bisector of $$\angle \mathrm{AOB}$$ meets the line $$\mathrm{AB}$$ at $$\mathrm{C}$$, then the length of $$OC$$ is

Let $$PQR$$ be a triangle with $$R(-1,4,2)$$. Suppose $$M(2,1,2)$$ is the mid point of $$\mathrm{PQ}$$. The distance of the centroid of $$△\mathrm{PQR}$$ from the point of intersection of the lines $$\frac{x-2}{0}=\frac{y}{2}=\frac{z+3}{-1}$$ and $$\frac{x-1}{1}=\frac{y+3}{-3}=\frac{z+1}{1}$$ is

Let $$\mathrm{P}(3,2,3),\mathrm{Q}(4,6,2)$$ and $$\mathrm{R}(7,3,2)$$ be the vertices of $$△\mathrm{PQR}$$. Then, the angle $$\angle \mathrm{QPR}$$ is

Let $$L_1:\vec{r}=(\hat{i}-\hat{j}+2\hat{k})+\lambda (\hat{i}-\hat{j}+2\hat{k}),\lambda \in \mathbb{R}$$, $$L_2:\vec{r}=(\hat{j}-\hat{k})+\mu (3\hat{i}+\hat{j}+p\hat{k}),\mu \in \mathbb{R}\text{, and }{L}_3:\vec{r}=\delta (\ell \hat{i}+m\hat{j}+n\hat{k}),\delta \in \mathbb{R}$$ be three lines such that $$L_1$$ is perpendicular to $$L_2$$ and $$L_3$$ is perpendicular to both $$L_1$$ and $$L_2$$. Then, the point which lies on $$L_3$$ is

Let $$(\alpha ,\beta ,\gamma )$$ be the foot of perpendicular from the point $$(1,2,3)$$ on the line $$\frac{x+3}{5}=\frac{y-1}{2}=\frac{z+4}{3}$$. Then $$19(\alpha +\beta +\gamma )$$ is equal to :

Let $$A(2,3,5)$$ and $$C(-3,4,-2)$$ be opposite vertices of a parallelogram $$ABCD$$. If the diagonal $$\overrightarrow{\mathrm{BD}}=\hat{i}+2\hat{j}+3\hat{k}$$, then the area of the parallelogram is equal to :

Consider the line $$\mathrm{L}$$ passing through the points $$(1,2,3)$$ and $$(2,3,5)$$. The distance of the point $$\left(\frac{11}{3},\frac{11}{3},\frac{19}{3}\right)$$ from the line $$\mathrm{L}$$ along the line $$\frac{3x-11}{2}=\frac{3y-11}{1}=\frac{3z-19}{2}$$ is equal to

The shortest distance between the lines $$\frac{x-3}{4}=\frac{y+7}{-11}=\frac{z-1}{5}$$ and $$\frac{x-5}{3}=\frac{y-9}{-6}=\frac{z+2}{1}$$ is:

Let the line $$\mathrm{L}$$ intersect the lines $$x-2=-y=z-1,2(x+1)=2(y-1)=z+1$$ and be parallel to the line $$\frac{x-2}{3}=\frac{y-1}{1}=\frac{z-2}{2}$$. Then which of the following points lies on $$\mathrm{L}$$ ?

Let the point, on the line passing through the points $$P(1,-2,3)$$ and $$Q(5,-4,7)$$, farther from the origin and at a distance of 9 units from the point $$P$$, be $$(\alpha ,\beta ,\gamma )$$. Then $${\alpha }^2+{\beta }^2+{\gamma }^2$$ is equal to :

Let $$\mathrm{P}$$ be the point of intersection of the lines $$\frac{x-2}{1}=\frac{y-4}{5}=\frac{z-2}{1}$$ and $$\frac{x-3}{2}=\frac{y-2}{3}=\frac{z-3}{2}$$. Then, the shortest distance of $$\mathrm{P}$$ from the line $$4x=2y=z$$ is

If the shortest distance between the lines $$\frac{x-\lambda }{2}=\frac{y-4}{3}=\frac{z-3}{4}$$ and $$\frac{x-2}{4}=\frac{y-4}{6}=\frac{z-7}{8}$$ is $$\frac{13}{\sqrt{29}}$$, then a value of $$\lambda$$ is :