If a, b c are coplanar vectors, then which of the following is not correct ?

• $\mathrm{a} . \left(\mathrm{b} \times \mathrm{c}\right) = 0$

• $\mathrm{a} \times \left(\mathrm{b} \times \mathrm{c}\right) = 0$

• [a + b, b + c, c + a] = 0

• a = pb + qc

Find the equation of plane through the line $\frac{\mathrm{x} - 2}{2} = \frac{\mathrm{y} - 3}{3} = \frac{\mathrm{z} - 4}{5}$ and parallel to X-axis.

• 2x + 3y + 5z = 1

• 2x - 3z - 3 = 0

• 5y - 3z - 3 = 0

• 3y + 4z = 0

The value of $\text{'}\mathrm{\lambda }\text{'}$, so that the vectors $\hat{\mathrm{i}} - 3\hat{\mathrm{j}} + \hat{\mathrm{k}}$, $2\hat{\mathrm{i}} + \mathrm{\lambda }\hat{\mathrm{j}} + \hat{\mathrm{k}}$ and $3\hat{\mathrm{i}} + \hat{\mathrm{j}} - 2\hat{\mathrm{k}}$ are coplanar, will be

• 0

• 2

• $- \frac{1}{2}$

• - 4

The line passing through the point (- 1, 2 3) and perpendicular to the plane x - 2y + 3z+ 5 = 0 will be

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

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

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

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

The value of k, if the line $\frac{\mathrm{x} - 4}{1} = \frac{\mathrm{y} - 2}{1} = \frac{\mathrm{z} - \mathrm{k}}{1}$ lies on the plane 2x - 4y + z = 7, will be

• 5

• 7

• 9

• 11

If the line of intersection of the planes 2x + 3y + z = 1 and x + 3y + 2 = 2 makes angle $\mathrm{\alpha }$ with positive direction of x  axis, then $\cos \left(\mathrm{\alpha }\right)$ will be equal to

• $\frac{1}{\sqrt{2}}$

• $\frac{1}{\sqrt{5}}$

• $\frac{1}{\sqrt{7}}$

• $\frac{1}{\sqrt{3}}$

Value of a for which the vectors (2, - 1, 1) (1, 2, - 3) and (3, a, 5) become coplanar will be

• 4

• - 4

• no such exists

• None of these

If l , m, n are the DC's of a line, then

• l² + m² + n² = 0

• l² + m² + n² = 1

• l + m + n = 1

• l = m = n = 1

The length of the perpendicular from the point (1 2, 3) on the line $\frac{\mathrm{x} - 6}{3} = \frac{\mathrm{y} - 7}{2} = \frac{\mathrm{z} - 7}{- 2}$ is

• 3 units

• 4 units

• 5 units

• 7 units

690.

The equation of the plane passing through the intersection ofthe planes 2x - 3y + z - 4 = 0 and x - y + z + 1 = 0 and perpendicular to the plane x + 2y - 3z + 6 = 0 is

• x - 5y + 3z - 23 = 0

• x - 5y - 3z - 23 = 0

• x + 5y - 3z + 23 = 0

• x - 5y + 3z + 23 = 0