The plane passing through the points (1, 2, 1), (2, 1, 2) and parallel to the line, 2x = 3y, z = 1 also passes through the point :

• (2, 0, - 1)

• (0, 6, - 2)

• (0,  - 6, 2)

• (- 2, 0 , 1)

A plane passing through the point (3, 1, 1) contains two lines whose direction ratios are 1, – 2, 2 and 2, 3, – 1 respectively. If this plane also passes through the point($\mathrm{\alpha }$,–3, 5), then $\mathrm{\alpha }$ is equal to

• 5

• 10

•  - 5

•  - 10

Let the latus rectum of the parabola y² = 4x be the common chord to the circles C₁ and C₂ each of them having radius 25. Then, the distance between the centres of the circles C₁ and C₂ is :

• 12

• 8

• $8\sqrt{5}$

• $4\sqrt{5}$

$$\begin{gathered} \mathrm{Let} \mathrm{a} \mathrm{plane} \mathrm{P} \mathrm{contain} \mathrm{two} \mathrm{lines} \\ \stackrel{\rightharpoonup }{\mathrm{r}} = \hat{\mathrm{i}} + \mathrm{\lambda }\left(\hat{\mathrm{i}} + \hat{\mathrm{j}}\right), \mathrm{\lambda } \in \mathrm{R} \mathrm{and} \\ \overrightarrow{\mathrm{r}} = - \hat{\mathrm{j}} + \mathrm{\mu }\left(\hat{\mathrm{j}} - \hat{\mathrm{k}}\right), \mathrm{\mu } \in \mathrm{R} \\ \mathrm{If} \mathrm{Q}\left(\mathrm{\alpha }, \mathrm{\beta }, \mathrm{\gamma }\right) \mathrm{is} \mathrm{the} \mathrm{foot} \mathrm{of} \mathrm{the} \mathrm{perpendicular} \mathrm{drawn} \mathrm{from} \mathrm{the} \mathrm{point} \mathrm{equals}... \end{gathered}$$

The shortest distance between the lines $\frac{\mathrm{x} - 1}{0} = \frac{\mathrm{y} + 1}{ - 1} = \frac{\mathrm{z}}{1} \mathrm{and} \mathrm{x} + \mathrm{y} + \mathrm{z} + 1 = 0, 2\mathrm{x} - \mathrm{y} + \mathrm{z} + 3 = 0 \mathrm{is}$

• $\frac{1}{2}$

• 1

• $\frac{1}{\sqrt{3}}$

• $\frac{1}{\sqrt{2}}$

The angle of elevation of the top of a hill from a point on the horizontal plane passing through the foot of the hill is found to be 45°. After waling a distance of 80 meters towards the top, up a slope inclined at angle of 30° to the horizontal plane the angle of elevation of the top of the hill becomes 75°. Then the height of the hill (in meters) is _____.

A plane P meets the coordinate axes at A, B and C respectively. The centroid of $∆$ABC is given to be (1, 1, 2). Then the equation of the line through this centroid and perpendicular to the plane P is:

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

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

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

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