531.

The shortest distance from the point (1, 2, - 1) to the surface of the sphere x² + y² + z² = 24 is :

• $3\sqrt{6}$ unit

• $\sqrt{6}$ unit

• $2\sqrt{6}$

• 2 sq unit

The equation of the plane which bisects the line joining (2, 3, 4) and (6, 7, 8) is :

• x - y - z - 15 = 0

• x - y - z - 15 = 0

• x + y + z - 15 = 0

• x + y + z + 15 = 0

A line makes acute angles of $\mathrm{\alpha }, \mathrm{\beta } \mathrm{and} \mathrm{\gamma }$ with the co-ordinate axes such that $\cos \left(\mathrm{\alpha }\right)\cos \left(\mathrm{\beta }\right) = \cos \left(\mathrm{\beta }\right)\cos \left(\mathrm{\gamma }\right) = \frac{2}{9} \mathrm{and} \cos \left(\mathrm{\gamma }\right)\cos \left(\mathrm{\alpha }\right) = \frac{4}{9}$, then $\cos \left(\mathrm{\alpha }\right) + \cos \left(\mathrm{\beta }\right) + \cos \left(\mathrm{\gamma }\right)$ is equal to :

• $\frac{25}{9}$

• $\frac{5}{9}$

• $\frac{5}{3}$

• $\frac{2}{3}$

The equation of the plane through the point (1, 2, 3), (- 1,  4, 2) and (3, 1, 1) is :

• 5x + y + 12z - 23 = 0

• 5x + 6y + 2z - 23 = 0

• x + 6y + 2z - 13 = 0

• x + y + z - 13 = 0

The number of solutions of the equation tan(x) + sec(x) = 2cos(x) and cos(x) $\ne$ 0 lying in the interval ($0, \mathrm{\pi }$) is :

• 2

• 1

• 0

• 3

The angle between $\overrightarrow{\mathrm{a}} \mathrm{and} \overrightarrow{\mathrm{b}}$ is $\frac{5\mathrm{\pi }}{6}$ and the projection of $\overrightarrow{\mathrm{a}}$ in the direction of $\overrightarrow{\mathrm{b}}$ is $\frac{- 6}{\sqrt{3}}$, then $\left|\overrightarrow{\mathrm{a}}\right|$ is equal to :

• 6

• $\frac{\sqrt{3}}{2}$

• 12

• 4

A unit vector in the plane of $\hat{\mathrm{i}} + 2\hat{\mathrm{j}} + \hat{\mathrm{k}}$ and $\hat{\mathrm{i}} + \hat{\mathrm{j}} + 2\hat{\mathrm{k}}$ and perpendicular yo $2\hat{\mathrm{i}} + \hat{\mathrm{j}} + \hat{\mathrm{k}}$ is :

• $\hat{\mathrm{j}} - \hat{\mathrm{k}}$

• $\frac{\hat{\mathrm{i}} + \hat{\mathrm{j}}}{\sqrt{2}}$

• $\frac{\hat{\mathrm{j}} + \hat{\mathrm{k}}}{\sqrt{2}}$

• $\frac{\hat{\mathrm{j}} - \hat{\mathrm{k}}}{\sqrt{2}}$

The equation of the plane through the point (2, - 1, - 3) and parallel to the lines $\frac{\mathrm{x} - 1}{3} = \frac{\mathrm{y} + 2}{2} = \frac{\mathrm{z}}{- 4}$ and $\frac{\mathrm{x}}{2} = \frac{\mathrm{y} - 1}{- 3} = \frac{\mathrm{z} - 2}{2}$ is :

• 8x + 14y + 13z + 37 = 0

• 8x - 14y + 13z + 37 = 0

• 8x + 14y - 13z + 37 = 0

• 8x + 14y + 13z - 37 = 0

If for a plane, the intercepts on the co-ordinate axes are 8, 4, 4, then the length of the perpendicular from the origin on to the plane is :

• $\frac{8}{3}$

• $\frac{3}{8}$

• 3

• $\frac{4}{3}$

If a plane meets the co-ordinate axes at A, B and C such that the centroid of the triangle is (1, 2, 4), then the equation of the plane is:

• x + 2y + 4z = 12

• 4x + 2y + z = 12

• x + 2y + 4z = 3

• 4x + 2y + z = 3