If the lines $\frac{1 - \mathrm{x}}{3} = \frac{\mathrm{y} - 2}{2\mathrm{\alpha }} = \frac{\mathrm{z} - 3}{2}$$\frac{\mathrm{x} - 1}{3\mathrm{\alpha }} = \mathrm{y} - 1 = \frac{6 - \mathrm{z}}{5}$ are perpendicular, then the value of $\mathrm{\alpha }$ is

• $\frac{- 10}{7}$

• $\frac{10}{7}$

• $\frac{- 10}{11}$

• $\frac{10}{11}$

112.

The distance between the lines $\overrightarrow{\mathrm{r}} = \left(4\hat{\mathrm{i}} - 7\hat{\mathrm{j}} - 9\hat{\mathrm{k}}\right) + \mathrm{t}\left(3\hat{\mathrm{i}} - 7\hat{\mathrm{j}} + 4\hat{\mathrm{k}}\right)$ and $\overrightarrow{\mathrm{r}} = \left(7\hat{\mathrm{i}} - 14\hat{\mathrm{j}} - 5\hat{\mathrm{k}}\right) + \mathrm{s}\left(- 3\hat{\mathrm{i}} + 7\hat{\mathrm{j}} - 4\hat{\mathrm{k}}\right)$ is equal to

• 1

• $\frac{1}{2}$

• $\frac{3}{4}$

• 0

Let T_n denote the number of triangles which can equal to be formed by using the vertices of a regular polygon of n sides. If T_{n + 1} - T_n = 28, then n equals

• 4

• 5

• 6

• 8

A plane makes intercepts a, b, cat A, B, C on the coordinate axes respectively. If the centroid of the $∆$ABC is at (3, 2, 1), then the equation of the plane is

• x + 2y + 3z = 9

• 2x - 3y - 6z = 18

• 2x + 3y + 6z = 18

• 2x + y + 6z = 18

If the plane 3x + y + 2z + 6 = 0 is parallel to the line $\frac{3\mathrm{x} - 1}{2\mathrm{b}} = 3 - \mathrm{y} = \frac{\mathrm{z} - 1}{\mathrm{a}}$, then the avlue of 3a + 3b is

• $\frac{1}{2}$

• $\frac{3}{2}$

• 3

• 4

The equation of the line passing through the point (3, 0,- 4) and perpendicular to the plane 2x - 3y + 5z - 7 = 0 is

• $\frac{\mathrm{x} - 2}{3} = \frac{\mathrm{y}}{- 3} = \frac{\mathrm{z} + 4}{5}$

• $\frac{\mathrm{x} - 3}{2} = \frac{\mathrm{y}}{- 3} = \frac{\mathrm{z} - 4}{5}$

• $\frac{\mathrm{x} - 2}{3} = \frac{- \mathrm{y}}{3} = \frac{\mathrm{z} + 4}{5}$

• $\frac{\mathrm{x} + 3}{2} = \frac{\mathrm{y}}{3} = \frac{\mathrm{z} - 4}{5}$

The plane $\overrightarrow{\mathrm{r}} = \mathrm{s}\left(\hat{\mathrm{i}} + \hat{\mathrm{j}} - 4\hat{\mathrm{k}}\right)$ $+ \mathrm{t}\left(3\hat{\mathrm{i}} +4 \hat{\mathrm{j}} - 4\hat{\mathrm{k}}\right)$ + $\left(1 - \mathrm{t}\right)\left(2\hat{\mathrm{i}} - 7\hat{\mathrm{j}} - 3\hat{\mathrm{k}}\right)$ is parallel to the line

• $\overrightarrow{\mathrm{r}} = \left(- \hat{\mathrm{i}} + \hat{\mathrm{j}} - \hat{\mathrm{k}}\right) + \mathrm{t}\left(- \hat{\mathrm{i}} - 2\hat{\mathrm{j}} + 4\hat{\mathrm{k}}\right)$

• $\overrightarrow{\mathrm{r}} = \left(- \hat{\mathrm{i}} + \hat{\mathrm{j}} - \hat{\mathrm{k}}\right) + \mathrm{t}\left( \hat{\mathrm{i}} - 2\hat{\mathrm{j}} + 4\hat{\mathrm{k}}\right)$

• $\overrightarrow{\mathrm{r}} = \left(\hat{\mathrm{i}} + \hat{\mathrm{j}} - \hat{\mathrm{k}}\right) + \mathrm{t}\left(- \hat{\mathrm{i}} - 4\hat{\mathrm{j}} + 7\hat{\mathrm{k}}\right)$

• $\overrightarrow{\mathrm{r}} = \left(- \hat{\mathrm{i}} + \hat{\mathrm{j}} - 3\hat{\mathrm{k}}\right) + \mathrm{t}\left(2\hat{\mathrm{i}} + 6\hat{\mathrm{j}} - 8\hat{\mathrm{k}}\right)$

The distance between the line $\overrightarrow{\mathrm{r}} = \left(2\hat{\mathrm{i}} + 2\hat{\mathrm{j}} - \hat{\mathrm{k}}\right) + \mathrm{\lambda }\left(2\hat{\mathrm{i}} + \hat{\mathrm{j}} - 2\hat{\mathrm{k}}\right)$ and the plane $\overrightarrow{\mathrm{r}} . \left(\hat{\mathrm{i}} + 2\hat{\mathrm{j}} + 2\hat{\mathrm{k}}\right)$ = 10 is equal to

• 5

• 4

• 3

• 2

Equation of the plane passing through t intersection of the planes x + y + z = 6 and 2x + 3y + 4z + 5 = 0 and the point (1, 1, 1)

• 20x + 23y + 26z - 69 = 0

• 31x + 45y + 49z + 52 = 0

• 8x + 5y + 2z - 69 = 0

• 4x + 5y + 6z - 7 = 0

The equation of the plane containing the line $\frac{\mathrm{x} - 1}{2}$ = $\frac{\mathrm{y} + 1}{- 1}$ = $\frac{\mathrm{z}}{3}$ and $\frac{\mathrm{x}}{2}$ = $\frac{\mathrm{y} - 2}{- 1}$ = $\frac{\mathrm{z} + 1}{3}$ is

• 8x - y + 5z - 8 = 0

• 8x + y - 5z - 7 = 0

• x - 8y + 3z + 6 = 0

• 8x + y - 5z + 7 = 0