Let the foot of the perpendicular from the point (1, 2, 4) on the line {{x + 2} \over 4} = {{y - 1} \over 2} = {{z + 1} \over 3} be P. Then the distance of P from the plane $$3x+4y+12z+23=0$$ is :

The shortest distance between the lines {{x - 3} \over 2} = {{y - 2} \over 3} = {{z - 1} \over { - 1}} and {{x + 3} \over 2} = {{y - 6} \over 1} = {{z - 5} \over 3}, is :

If two straight lines whose direction cosines are given by the relations $$l+m-n=0$$, $$3l^2+m^2+cnl=0$$ are parallel, then the positive value of c is :

If the two lines {l_1}:{{x - 2} \over 3} = {{y + 1} \over {-2}},\,z = 2 and {l_2}:{{x - 1} \over 1} = {{2y + 3} \over \alpha } = {{z + 5} \over 2} are perpendicular, then an angle between the lines l2 and {l_3}:{{1 - x} \over 3} = {{2y - 1} \over { - 4}} = {z \over 4} is :

Let the plane 2x + 3y + z + 20 = 0 be rotated through a right angle about its line of intersection with the plane x $$-$$ 3y + 5z = 8. If the mirror image of the point \left( {2, - {1 \over 2},2} \right) in the rotated plane is B(a, b, c), then :

If the plane $$2x+y-5z=0$$ is rotated about its line of intersection with the plane $$3x-y+4z-7=0$$ by an angle of {\pi \over 2}, then the plane after the rotation passes through the point :

If the lines $$\overrightarrow{r}=\left(\hat{i}-\hat{j}+\hat{k}\right)+\lambda \left(3\hat{j}-\hat{k}\right)$$ and $$\overrightarrow{r}=\left(\alpha \hat{i}-\hat{j}\right)+\mu \left(2\hat{i}-3\hat{k}\right)$$ are co-planar, then the distance of the plane containing these two lines from the point ($$\alpha$$, 0, 0) is :

Let $$\overrightarrow{a}=\hat{i}+\hat{j}+2\hat{k}$$, $$\overrightarrow{b}=2\hat{i}-3\hat{j}+\hat{k}$$ and $$\overrightarrow{c}=\hat{i}-\hat{j}+\hat{k}$$ be three given vectors. Let $$\overrightarrow{v}$$ be a vector in the plane of $$\overrightarrow{a}$$ and $$\overrightarrow{b}$$ whose projection on $$\overrightarrow{c}$$ is {2 \over {\sqrt 3 }}. If $$\overrightarrow{v}.\hat{j}=7$$, then $$\overrightarrow{v}.\left(\hat{i}+\hat{k}\right)$$ is equal to :

Let p be the plane passing through the intersection of the planes $$\overrightarrow{r}.\left(\hat{i}+3\hat{j}-\hat{k}\right)=5$$ and $$\overrightarrow{r}.\left(2\hat{i}-\hat{j}+\hat{k}\right)=3$$, and the point (2, 1, $$-$$2). Let the position vectors of the points X and Y be $$\hat{i}-2\hat{j}+4\hat{k}$$ and $$5\hat{i}-\hat{j}+2\hat{k}$$ respectively. Then the points :

Let Q be the mirror image of the point P(1, 0, 1) with respect to the plane S : x + y + z = 5. If a line L passing through (1, $$-$$1, $$-$$1), parallel to the line PQ meets the plane S at R, then QR2 is equal to :

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

Let the points on the plane P be equidistant from the points ($$-$$4, 2, 1) and (2, $$-$$2, 3). Then the acute angle between the plane P and the plane 2x + y + 3z = 1 is :

The distance of the point (3, 2, $$-$$1) from the plane $$3x-y+4z+1=0$$ along the line {{2 - x} \over 2} = {{y - 3} \over 2} = {{z + 1} \over 1} is equal to :

Let $$\mathrm{P}$$ be the plane containing the straight line $$\frac{x-3}{9}=\frac{y+4}{-1}=\frac{z-7}{-5}$$ and perpendicular to the plane containing the straight lines $$\frac{x}{2}=\frac{y}{3}=\frac{z}{5}$$ and $$\frac{x}{3}=\frac{y}{7}=\frac{z}{8}$$. If $$\mathrm{d}$$ is the distance of $$\mathrm{P}$$ from the point $$(2,-5,11)$$, then $${\mathrm{d}}^2$$ is equal to :

A plane $$E$$ is perpendicular to the two planes $$2x-2y+z=0$$ and $$x-y+2z=4$$, and passes through the point $$P(1,-1,1)$$. If the distance of the plane $$E$$ from the point $$Q(a,a,2)$$ is $$3\sqrt{2}$$, then $$(PQ{)}^2$$ is equal to :

The shortest distance between the lines $$\frac{x+7}{-6}=\frac{y-6}{7}=z$$ and $$\frac{7-x}{2}=y-2=z-6$$ is :

The length of the perpendicular from the point $$(1,-2,5)$$ on the line passing through $$(1,2,4)$$ and parallel to the line $$x+y-z=0=x-2y+3z-5$$ is :

A vector $$\vec{a}$$ is parallel to the line of intersection of the plane determined by the vectors $$\hat{i},\hat{i}+\hat{j}$$ and the plane determined by the vectors $$\hat{i}-\hat{j},\hat{i}+\hat{k}$$. The obtuse angle between $$\vec{a}$$ and the vector $$\vec{b}=\hat{i}-2\hat{j}+2\hat{k}$$ is :

If the plane $$P$$ passes through the intersection of two mutually perpendicular planes $$2x+ky-5z=1$$ and $$3 k x-k y+z=5, k

If the length of the perpendicular drawn from the point $$P(a,4,2)$$, a $$>0$$ on the line $$\frac{x+1}{2}=\frac{y-3}{3}=\frac{z-1}{-1}$$ is $$2\sqrt{6}$$ units and $$Q\left({\alpha }_1,{\alpha }_2,{\alpha }_3\right)$$ is the image of the point P in this line, then $$\mathrm{a}+\sum _{i=1}^3{\alpha }_i$$ is equal to :