$$\begin{gathered} \mathrm{The} \mathrm{distance} \mathrm{of} \mathrm{the} \mathrm{point} (1, – 2, 3) \mathrm{from} \mathrm{the} \mathrm{plane} \mathrm{x} – \mathrm{y} + \mathrm{z} = 5 \\ \mathrm{measured} \mathrm{parallel} \mathrm{to} \mathrm{the} \mathrm{line} \frac{\mathrm{x}}{2} = \frac{\mathrm{y}}{3} = \frac{\mathrm{z}}{ - 6} \mathrm{is} : \end{gathered}$$

• $\frac{1}{7}$

• 7

• 1

• $\frac{7}{5}$

If the perpendicular bisector of the line segment joining the points P(1, 4) and Q(k, 3) has y-intercept equal to – 4, then a value of k is ;

• $\sqrt{15}$

•  4

•  - 2

• $\sqrt{14}$

213.

If (a, b, c) is the image of the point (1, 2, – 3) in the line,

$\frac{\mathrm{x} + 1}{2} = \frac{\mathrm{y} - 3}{ - 2} = \frac{\mathrm{z}}{ - 1}$, then a + b + c is equal to :

• 3

• - 1

• 2

• 1

If the line, 2x – y + 3 = 0 is at a distance $\frac{1}{\sqrt{5}} \mathrm{and} \frac{2}{\sqrt{5}} \mathrm{from} \mathrm{the} \mathrm{lines} 4\mathrm{x} - 2\mathrm{y} + \mathrm{\alpha } = 0 \mathrm{and} 6\mathrm{x} - 3\mathrm{y} + \mathrm{\beta } = 0,$

hen the sum of all possible values of $\mathrm{\alpha } \mathrm{and} \mathrm{\beta }$ is_

$$\begin{gathered} \mathrm{If} \mathrm{for} \mathrm{some} \mathrm{\alpha } \in \mathrm{R}, \mathrm{the} \mathrm{lines} \\ {\mathrm{L}}_1 : \frac{\mathrm{x} + 1}{2} = \frac{\mathrm{y} - 2}{ - 1} = \frac{\mathrm{z} - 1}{1} \mathrm{and} \\ {\mathrm{L}}_2 :\hspace{0.17em}\frac{\mathrm{x} + 2}{\mathrm{\alpha }} = \frac{\mathrm{y} + 1}{5 - \mathrm{\alpha }} = \frac{\mathrm{z} + 1}{1} \mathrm{are} \mathrm{coplanar}, \mathrm{then} \mathrm{the} \mathrm{line} {\mathrm{L}}_2 \mathrm{passes} \mathrm{through} \mathrm{point} : \end{gathered}$$

• (2, - 10, - 2)

• (10, 2, 2)

• ( - 2, 10, 2)

• (10, - 2, - 2)

Which of the following points lies on the tangent to the curve${\mathrm{x}}^4{\mathrm{e}}^{\mathrm{y}} + 2\sqrt{\mathrm{y} + 1} = 3$ at the point (1, 0)

• (- 2, 4)

• (2, 2)

• ( - 2, 6)

• (2, 6)

A ray of light coming from the point (2, 2$\sqrt{3}$) is incident at an angle 30° on the line x = 1 at the point A. The ray gets reflected on the line x = 1 and meets x-axis at the point B. Then, the line AB passes through the point ;

• $\left(4, - \frac{\sqrt{3}}{2}\right)$

• $\left(3, - \sqrt{3}\right)$

• $(4, - \sqrt{3})$

• $\left(3, - \frac{1}{\sqrt{3}}\right)$

Let L denote the line in the xy-plane with x and y intercepts as 3 respectively. Then the image of the point (– 1, – 4) in the line is :

• $\left(\frac{8}{5}, \frac{29}{5}\right)$

• $\left(\frac{11}{5}, \frac{28}{5}\right)$

• $\left(\frac{29}{5}, \frac{8}{5}\right)$

• $\left(\frac{29}{5}, \frac{11}{5}\right)$

If the pair of straight lines xy - x - y + 1 = 0 and the line ax + 2y - 3a = 0 are concurrent, then a is equal to

• 0

• 1

• - 1

• 3

The mid-point of the line segment joining the centroid and the orthocentre of the triangle whose vertices are (a, b),(a, c) and (d, c), is

• $\left(\frac{5\mathrm{a} +\hspace{0.17em}\mathrm{d}}{6}, \frac{\mathrm{b} + 5\mathrm{c}}{6}\right)$

• `$\left(\frac{\mathrm{a} + 5\mathrm{d}}{6}, \frac{5\mathrm{b} + \mathrm{c}}{6}\right)$

• (a, 0)

• (0, 0)