The shortest distance between the lines $$x+1=2y=-12z$$ and $$x=y+2=6z-6$$ is :

The distance of the point P(4, 6, $$-$$2) from the line passing through the point ($$-$$3, 2, 3) and parallel to a line with direction ratios 3, 3, $$-$$1 is equal to :

Consider the lines $$L_1$$ and $$L_2$$ given by {L_1}:{{x - 1} \over 2} = {{y - 3} \over 1} = {{z - 2} \over 2} {L_2}:{{x - 2} \over 1} = {{y - 2} \over 2} = {{z - 3} \over 3}. A line $$L_3$$ having direction ratios 1, $$-$$1, $$-$$2, intersects $$L_1$$ and $$L_2$$ at the points $$P$$ and $$Q$$ respectively. Then the length of line segment $$PQ$$ is

If the foot of the perpendicular drawn from (1, 9, 7) to the line passing through the point (3, 2, 1) and parallel to the planes $$x+2y+z=0$$ and $$3y-z=3$$ is ($$\alpha ,\beta ,\gamma$$), then $$\alpha +\beta +\gamma$$ is equal to :

Let the plane containing the line of intersection of the planes P1 : $$x+(\lambda +4)y+z=1$$ and P2 : $$2x+y+z=2$$ pass through the points (0, 1, 0) and (1, 0, 1). Then the distance of the point (2$$\lambda ,\lambda ,-\lambda$$) from the plane P2 is :

The distance of the point (7, $$-$$3, $$-$$4) from the plane passing through the points (2, $$-$$3, 1), ($$-$$1, 1, $$-$$2) and (3, $$-$$4, 2) is :

The distance of the point ($$-1,9,-16$$) from the plane $$2x+3y-z=5$$ measured parallel to the line {{x + 4} \over 3} = {{2 - y} \over 4} = {{z - 3} \over {12}} is :

Let the foot of perpendicular of the point $P(3,-2,-9)$ on the plane passing through the points $(-1,-2,-3),(9,3,4),(9,-2,1)$ be $Q(\alpha ,\beta ,\gamma )$. Then the distance of $Q$ from the origin is :

Let the system of linear equations $-x+2 y-9 z=7$ $-x+3 y+7 z=9$ $-2 x+y+5 z=8$ $-3x+y+13z=\lambda$ has a unique solution $x=\alpha ,y=\beta ,z=\gamma$. Then the distance of the point $(\alpha ,\beta ,\gamma )$ from the plane $2x-2y+z=\lambda$ is :

Let $\mathrm{S}$ be the set of all values of $\lambda$, for which the shortest distance between the lines $\frac{x-\lambda }{0}=\frac{y-3}{4}=\frac{z+6}{1}$ and $\frac{x+\lambda }{3}=\frac{y}{-4}=\frac{z-6}{0}$ is 13. Then $8\left|\sum _{\lambda \in S}\lambda \right|$ is equal to :

The line, that is coplanar to the line $$\frac{x+3}{-3}=\frac{y-1}{1}=\frac{z-5}{5}$$, is :

The plane, passing through the points $$(0,-1,2)$$ and $$(-1,2,1)$$ and parallel to the line passing through $$(5,1,-7)$$ and $$(1,-1,-1)$$, also passes through the point :

Let $$\mathrm{N}$$ be the foot of perpendicular from the point $$\mathrm{P}(1,-2,3)$$ on the line passing through the points $$(4,5,8)$$ and $$(1,-7,5)$$. Then the distance of $$N$$ from the plane $$2x-2y+z+5=0$$ is :

Let the equation of plane passing through the line of intersection of the planes $$x+2y+az=2$$ and $$x-y+z=3$$ be $$5x-11y+bz=6a-1$$. For $$c\in \mathbb{Z}$$, if the distance of this plane from the point $$(a,-c,c)$$ is $$\frac{2}{\sqrt{a}}$$, then $$\frac{a+b}{c}$$ is equal to :

The distance of the point $$(-1,2,3)$$ from the plane $$\vec{r}\cdot (\hat{i}-2\hat{j}+3\hat{k})=10$$ parallel to the line of the shortest distance between the lines $$\vec{r}=(\hat{i}-\hat{j})+\lambda (2\hat{i}+\hat{k})$$ and $$\vec{r}=(2\hat{i}-\hat{j})+\mu (\hat{i}-\hat{j}+\hat{k})$$ is :

Let the lines $$l_1:\frac{x+5}{3}=\frac{y+4}{1}=\frac{z-\alpha }{-2}$$ and $$l_2:3x+2y+z-2=0=x-3y+2z-13$$ be coplanar. If the point $$\mathrm{P}(a,b,c)$$ on $$l_1$$ is nearest to the point $$\mathrm{Q}(-4,-3,2)$$, then $$|a|+|b|+|c|$$ is equal to

Let the plane P: $$4x-y+z=10$$ be rotated by an angle $$\frac{\pi }{2}$$ about its line of intersection with the plane $$x+y-z=4$$. If $$\alpha$$ is the distance of the point $$(2,3,-4)$$ from the new position of the plane $$\mathrm{P}$$, then $$35\alpha$$ is equal to :

Let the line passing through the points $$\mathrm{P}(2,-1,2)$$ and $$\mathrm{Q}(5,3,4)$$ meet the plane $$x-y+z=4$$ at the point $$\mathrm{R}$$. Then the distance of the point $$\mathrm{R}$$ from the plane $$x+2y+3z+2=0$$ measured parallel to the line $$\frac{x-7}{2}=\frac{y+3}{2}=\frac{z-2}{1}$$ is equal to :

Let P be the plane passing through the points $$(5,3,0),(13,3,-2)$$ and $$(1,6,2)$$. For $$\alpha \in \mathbb{N}$$, if the distances of the points $$\mathrm{A}(3,4,\alpha )$$ and $$\mathrm{B}(2,\alpha ,a)$$ from the plane P are 2 and 3 respectively, then the positive value of a is :

Let $$(\alpha ,\beta ,\gamma )$$ be the image of the point $$\mathrm{P}(2,3,5)$$ in the plane $$2x+y-3z=6$$. Then $$\alpha +\beta +\gamma$$ is equal to :