If the planes $\overrightarrow{\mathrm{r}} . \left(2\hat{\mathrm{i}} - \mathrm{\lambda }\hat{\mathrm{j}} + 3\hat{\mathrm{k}}\right) = 0 \mathrm{and} \overrightarrow{\mathrm{r}} . \left(\mathrm{\lambda }\hat{\mathrm{i}} + 5\hat{\mathrm{j}} - \hat{\mathrm{k}}\right) = 5$  are perpendicular to each other, then the value of ${\mathrm{\lambda }}^2 + \mathrm{\lambda }$ is

• 0

• 2

• 1

• 3

The cartesian form of the plane $\overrightarrow{\mathrm{r}} = \left(\mathrm{s} - 2\mathrm{t}\right)\hat{\mathrm{i}} + \left(3 - \mathrm{t}\right)\hat{\mathrm{j}} + \left(2\mathrm{s} + \mathrm{t}\right)\hat{\mathrm{k}}$ is

• 2x - 5y -  z - 15 = 0

• 2x - 5y +  z - 15 = 0

• 2x - 5y -  z + 15 = 0

• 2x + 5y -  z + 15 = 0

Let P(- 7, 1, - 5) be a point on a plane and let O be the origin. If OP is normal to the plane, then the equation of the plane is

• 7x - y + 5z + 75 = 0

• 7x + y - 5z + 73 = 0

• 7x + y + 5z + 73 = 0

• 7x - y - 5z + 75 = 0

The shortest distance from the plane 12x + 4y + 3z = 327 to the sphere x² + y² + z² + 4x - 2y - 6z = 155 is

• 26

• $11\frac{4}{13}$

• 13

• 39

The point in the xy-plane which is equidistant from the point (2, 0, 3), (0, 3, 2) and (0, 0, 1) is

• (1, 2, 3)

• (- 3, 2, 0)

• (3, - 2, 0)

• (3, 2, 0)

The angle between the line $\frac{3\mathrm{x} - 1}{3} = \frac{\mathrm{y} +\hspace{0.17em}3}{- 1} = \frac{5 - 2\mathrm{z}}{4}$ and the plane 3x - 3y - 6z = 10 is equal to

• $\frac{\mathrm{\pi }}{6}$

• $\frac{\mathrm{\pi }}{4}$

• $\frac{\mathrm{\pi }}{3}$

• $\frac{\mathrm{\pi }}{2}$

The angle between the straight lines $\overrightarrow{\mathrm{r}} =\left(2 - 3\mathrm{t}\right)\hat{\mathrm{i}} + \left(1 + 2\mathrm{t}\right)\hat{\mathrm{j}} + \left(2 + 6\mathrm{t}\right)\hat{\mathrm{k}}$ and $\overrightarrow{\mathrm{r}} =\left(1 + 4\mathrm{s}\right)\hat{\mathrm{i}} + \left(2 - s\right)\hat{\mathrm{j}} + \left(8s - 1\right)\hat{\mathrm{k}}$ is

• ${\cos }^{-1}\left(\frac{\sqrt{41}}{34}\right)$

• ${\cos }^{-1}\left(\frac{21}{34}\right)$

• ${\cos }^{-1}\left(\frac{43}{63}\right)$

• ${\cos }^{-1}\left(\frac{34}{63}\right)$

If Q is the image of the point P(2, 3, 4) under the reflection in the plane x - 2y + 5z = 6, then the equation of the line PQ is

• $\frac{\mathrm{x} - 2}{- 1} = \frac{\mathrm{y} - 3}{2} = \frac{\mathrm{z} - 4}{5}$

• $\frac{\mathrm{x} - 2}{1} = \frac{\mathrm{y} - 3}{- 2} = \frac{\mathrm{z} - 4}{5}$

• $\frac{\mathrm{x} - 2}{- 1} = \frac{\mathrm{y} - 3}{- 2} = \frac{\mathrm{z} - 4}{5}$

• $\frac{\mathrm{x} - 2}{1} = \frac{\mathrm{y} - 3}{2} = \frac{\mathrm{z} - 4}{5}$

The distance of the point of intersection of the line $\frac{\mathrm{x} - 2}{3} = \frac{\mathrm{y} + 1}{4} = \frac{\mathrm{z} - 2}{12}$ and the plane x - y + z = 5 from the point (- 1, - 5, - 10)is

• 13

• 12

• 11

• 8

If the direction cosines of a line are $\left(\frac{1}{\mathrm{c}}, \frac{1}{\mathrm{c}}, \frac{1}{\mathrm{c}}\right)$, then

• 0 < c < 1

• c > 2

• $\mathrm{c} = \pm \sqrt{2}$

• $\mathrm{c} = \pm \sqrt{3}$