$$\begin{gathered} \mathrm{If} \mathrm{a} \mathrm{line} \mathrm{makes} \mathrm{angles} \mathrm{\alpha }, \mathrm{\beta }, \mathrm{\gamma } \mathrm{and} \mathrm{\delta } \mathrm{with} \mathrm{the} \mathrm{four} \mathrm{diagonals} \mathrm{of} \mathrm{a} \mathrm{cube}, \mathrm{then} \mathrm{the} \mathrm{value} \\ \mathrm{of} {\sin }^2\left(\mathrm{\alpha }\right) +{\sin }^2\left(\mathrm{\beta }\right) + {\sin }^2\left(\mathrm{\gamma }\right) + {\sin }^2\left(\mathrm{\delta }\right) \mathrm{is} \end{gathered}$$

• $\frac{4}{3}$

• $\frac{8}{3}$

• $\frac{7}{3}$

• $\frac{5}{3}$

If the pair of lines x - 16pxy - y² = 0 and x² - 16qxy - y = 0 are such that each pair bisects the angle between the other pair, then pq = 

• $\frac{- 1}{64}$

• $\frac{1}{64}$

• $\frac{ - 1}{8}$

• $\frac{1}{8}$

The equation of the pair of lines through the point (2, 1) and perpendicular to the pair of lines 4xy + 2x + 6y + 3 = 0 is

• xy - x - 2y + 2 = 0

• xy + x - 2y - 2 = 0

• xy + x + 2y - 6 = 0

• xy - x + 2y - 2 = 0

$$\begin{gathered} \mathrm{In} ∆\mathrm{ABC}, \mathrm{L}, \mathrm{M}, \mathrm{N} \mathrm{are} \mathrm{points} \mathrm{on} \mathrm{BC}, \mathrm{CA}, \mathrm{AB} \\ \mathrm{respectivley}, \mathrm{dividing} \mathrm{them} \mathrm{in} \mathrm{the} \mathrm{ratio} 1 : 2, \\ 2 : 3, 3: 5. \mathrm{If} \mathrm{the} \mathrm{point} \mathrm{K} \mathrm{divides} \mathrm{AB} \mathrm{in} \mathrm{the} \mathrm{ratio} 5: 3, \mathrm{then}\frac{\left|\overline{\mathrm{AL}} + \overline{\mathrm{BM}} + \overline{\mathrm{CN}}\right|}{\left|\overline{\mathrm{CK}}\right|} = ? \end{gathered}$$

• $\frac{5}{8}$

• $\frac{2}{5}$

• $\frac{3}{5}$

• $\frac{1}{15}$

The point to which the origin is to be shifted to remove the first degree terms from the equation 2x² + 4xy - 6y² + 2x + 8y + 1 = 0 is

• $\left(\frac{7}{8}, \frac{3}{8}\right)$

• $\left(\frac{- 7}{8}, \frac{ - 3}{8}\right)$

• $\left(\frac{- 7}{8}, \frac{3}{8}\right)$

• $\left(\frac{7}{8}, \frac{ - 3}{8}\right)$

The figure formed by the pairs of lines
$6{\mathrm{x}}^2 + 13\mathrm{xy} + 6{\mathrm{y}}^2 = 0$ and
6x² + 13xy + 6y + 10x + 10y + 4 = 0, is a

• Square

• Parallelogram

• Rhombus

• Rectangle

Let P(h, k) be a point on the curve y = x² + 7x + 2, nearest to the line, y = 3x – 3. Then the equation of the normal to the curve at P is :

• x + 3y - 62 = 0

• x + 3y + 26 = 0 

• x - 3y - 11 = 0

• x - 3y + 22 = 0

The Plane which bisects the line joining the points (4, – 2, 3) and (2, 4, – 1) at right angles also passes through the point :

• (0, - 1, 1)

• (4, 0, - 1)

• (4, 0, 1)

• (0, 1, - 1)

$\mathrm{If} \mathrm{a} ∆\mathrm{ABC} \mathrm{has} \mathrm{vertices} \mathrm{A}\left( - 1, 7\right), \mathrm{B}\left( - 7, 1\right) \mathrm{and} \mathrm{C}\left(5, - 5\right), \mathrm{then} \mathrm{its} \mathrm{orthocentre} \mathrm{has} \mathrm{coordinates}$

• ( - 3, 3)

• (3, - 3)

• $\left(- \frac{3}{5}, \frac{3}{5}\right)$

• $\left(\frac{3}{5}, - \frac{3}{5}\right)$

The angle of elevation of a cloud C from a point P, 200 m above a still take is 30°. If the angle of depression of the image of C in the lake from the point P is 60°, then PC (in m) is equal to

• 100

• $400\sqrt{3}$

• $200\sqrt{3}$

• 400