The distance between the lines $\frac{\mathrm{x} - 1}{3} = \frac{\mathrm{y} + 2}{- 2} = \frac{\mathrm{z} - 1}{2}$ and the plane 2x + 2y - z = 6 is

• 9

• 3

• 2

• 1

The cosine of the angle between any two diagonals of a cube is

• $\frac{1}{3}$

• $\frac{2}{3}$

• $- \frac{2}{3}$

• $\frac{1}{2}$

ABCD is a parallelogram, with AC, BD as diagonals, then AC - BD is equal to

• 4AB

• AB

• 3AB

• 2AB

If $\mathrm{\theta }$ is the angle between a and b and $\left|\mathrm{a} \times \mathrm{b}\right| = \left|\mathrm{a} . \mathrm{b}\right|$, then $\mathrm{\theta }$ is equal to

• 0

• $\mathrm{\pi }$

• $\frac{\mathrm{\pi }}{2}$

• $\frac{\mathrm{\pi }}{4}$

a, b, c, d are coplanar vectors, then (a x b) x (c x d) is equal to

• 0

• 1

• a

• b

If the foot of the perpendicular from (0, 0, 0) to the plane is (1, 2, 2), then the equation ofthe plane is

• - x + 2y + 8z - 9 = 0

• x + 2y + 2z - 9 = 0

• x + y + z - 5 = 0

• x + 2y - 3z + 1 = 0

If P = (0, 1, 2), Q =(4, - 2, 1), O =(0, 0, 0), then $\angle$POQ is equal to

• $\frac{\mathrm{\pi }}{2}$

• $\frac{\mathrm{\pi }}{4}$

• $\frac{\mathrm{\pi }}{6}$

• $\frac{\mathrm{\pi }}{3}$

A variable plane is at a constant distance h from the origin and meets the coordinate axes in A, B, C. Locus of centroid of ABC is

• x² + y² + z² = h^{- 2}

• x² + y² + z² = 4h^{- 2}

• x² + y² + z² = 16h²

• $\frac{1}{{\mathrm{x}}^2} + \frac{1}{{\mathrm{y}}^2} + \frac{1}{{\mathrm{z}}^2} = \frac{9}{{\mathrm{h}}^2}$

The acute angle between the two lines whose direction ratios are given by l + m - n = 0 and  ${\mathrm{l}}^2 + {\mathrm{m}}^2 + {\mathrm{n}}^2 = 0, \mathrm{is}$

• 0

• $\frac{\mathrm{\pi }}{6}$

• $\frac{\mathrm{\pi }}{4}$

• $\frac{\mathrm{\pi }}{3}$

The direction ratios of normal to the plane passing through (0, 0, 1), (0, 1, 2) and (1, 0, 3) are

• (2, 1, - 1)

• (1, 0, 1)

• (0, 0, - 1)

• (1, 0, 0)