161.

The plane x + 3y + 13 = 0 passes through the line of intersection of the planes 2x - 8y + 4z = p and 3x - 5y + 4z + 10 = 0. If the plane is perpendicular to the plane 3x - y - 2z - 4 = 0, then the value of p is

• 2

• 5

• 9

• 3

If a straight line makes the angles 60°, 45° and $\mathrm{\alpha }$ with X, Y and Z-axes respectively, then ${\sin }^2\left(\mathrm{\alpha }\right)$ is equal to

• $\frac{3}{4}$

• $\frac{3}{2}$

• $\frac{1}{2}$

• 1

The angle between a and b is $\frac{5\mathrm{\pi }}{6}$ and  and the projection of a on b is $\frac{- 9}{\sqrt{3}}$, then $\left|\mathrm{a}\right|$ is equal to

• 12

• 8

• 10

• 6

The direction cosines of the straight line given by the planes x = 0 and z = 0 are

• 1, 0, 0

• 0, 0, 1

• 1, 1, 0

• 0, 1, 0

If a . b = 0 and a + b makes an angle of 60° with b, then $\left|\mathrm{a}\right|$ is equal to

• 0

• $\frac{1}{\sqrt{3}}\left|\mathrm{b}\right|$

• $\frac{1}{\left|\mathrm{b}\right|}$

• $\sqrt{3}\left|\mathrm{b}\right|$

The straight line r = $(\hat{\mathrm{i}} + \hat{\mathrm{j}} + \hat{\mathrm{k}}) + \mathrm{a}(2\hat{\mathrm{i}} - \hat{\mathrm{j}} + 4\hat{\mathrm{k}})$ meets the XY - plane at the point

• (2, - 1, 0)

• (3, 4, 0)

• $\left(\frac{1}{2}, \frac{3}{4}, 0\right)$

• $\left(\frac{1}{2}, \frac{5}{4}, 0\right)$

The equation of the plane passing through (- 1, 5, - 7) and parallel to the plane 2x - 5y + 7z + 11 = 0, is

• $\mathrm{r}. \left(2\hat{\mathrm{i}} - 5\hat{\mathrm{j}} - 7\hat{\mathrm{k}}\right) + 76 = 0$

• $\mathrm{r}. \left(2\hat{\mathrm{i}} - 5\hat{\mathrm{j}} + 7\hat{\mathrm{k}}\right) + 76 = 0$

• $\mathrm{r}. \left(2\hat{\mathrm{i}} - 5\hat{\mathrm{j}} -+ 7\hat{\mathrm{k}}\right) + 75 = 0$

• $\mathrm{r}. \left(2\hat{\mathrm{i}} - 5\hat{\mathrm{j}} + 7\hat{\mathrm{k}}\right) + 65 = 0$

The angle subtended at the point (1, 2, 3) by the points P(2, 4, 5) and Q(3, 3, 1) is

• 90°

• 60°

• 30°

• 0°

If the two lines $\frac{\mathrm{x} - 1}{2} = \frac{1 - \mathrm{y}}{- \mathrm{a}} = \frac{\mathrm{z}}{4}$ and $\frac{\mathrm{x} - 3}{1} = \frac{2\mathrm{y} - 3}{4} = \frac{\mathrm{z} - 2}{2}$ are perpendicular, then the value of a is equal to

• - 4

• 5

• - 5

• 4

If the line $\frac{\mathrm{x} + 1}{2} = \frac{\mathrm{y} + 1}{3} = \frac{\mathrm{z} + 1}{4}$ meets the plane x + 2y + 3z = 14 at P, then the distance between P and the origin is

• $\sqrt{14}$

• $\sqrt{15}$

• $\sqrt{13}$

• $\sqrt{12}$