In the triangle with vertices at A(6, 3), B(- 6, 3) and C(- 6, - 3), the median through A meets BC at P, the line AC meets the x-axis at Q, while R and S respectively denote the orthocentre and centroid of the triangle. Then the correct matching of the coordinates of points in List-I to List-II is

      List-I                        List-II

(i)    P                      (A)    (0, 0)

(ii)   Q                      (B)    (6, 0)

(iii)   R                     (C)    (- 2, 1)

(iv)   S                     (D)    (- 6, 0)

                               (E)    (- 6, - 3)

                               (F)    (- 6, 3)

In $∆$ABC the mid points of the sides AB, BC and CA are respectively (l, 0, 0), (0, m, 0) and (0, 0, n). Then, $\frac{{\mathrm{AB}}^2 + {\mathrm{BC}}^2 + {\mathrm{CA}}^2}{{\mathrm{l}}^2 + {\mathrm{m}}^2 + {\mathrm{n}}^2} = ?$

• 2

• 4

• 8

• 16

$\mathrm{The} \mathrm{perimeter} \mathrm{of} \mathrm{the} \mathrm{triangle} \mathrm{with} \mathrm{vertices} \mathrm{at} \left(1, 0, 0\right), \left(0, 1, 0\right) \mathrm{and} \left(0, 0, 1\right) \mathrm{is}$

• 3

• 2

• $2\sqrt{2}$

• $3\sqrt{2}$

$$\begin{gathered} \mathrm{If} \mathrm{a} \mathrm{line} \mathrm{in} \mathrm{the} \mathrm{space} \mathrm{makes} \mathrm{angle} \mathrm{\alpha }, \mathrm{\beta } \mathrm{and} \mathrm{\gamma } \mathrm{with} \mathrm{the} \mathrm{coordinate} \mathrm{axes}, \mathrm{then} \\ \cos \left(2\mathrm{\alpha }\right) + \cos \left(2\mathrm{\beta }\right) + \cos \left(2\mathrm{\gamma }\right) + {\sin }^2\left(\mathrm{\alpha }\right) +\hspace{0.17em}{\sin }^2\left(\mathrm{\beta }\right) + {\sin }^2\left(\mathrm{\gamma }\right) =? \end{gathered}$$

• - 1

• 0

• 1

• 2

A plane meets the coordinate axes at A, B, C so that the centroid of the triangle ABC is (1, 2, 4). Then, the equation of the plane is

• x + 2y +4z =12

• 4x + 2y + z = 12

• x + 2y + 4z = 3

• 4x + 2y + z = 3

If (2, 3, - 3) is one end of a diameter of the sphere x² + y² + z² - 6x - 12y - 2z + 20 = 0, then the other end of the diameter is

• (4, 9, - 1)

• (4, 9, 5)

• (- 8, - 15, 1)

• (8, 15, 5)

The locus of a point such that the sum of its distances from the points (0, 2) and (0, - 2) is 6, is

• 9x² - 5y² = 45

• 5x² + 9y² = 45

• 9x² + 5y² = 45

• 5x² - 9y² = 45

The ratio in which the line joining (2, - 4, 3) and ( - 4, 5, - 6) is divided by the plane 3x + 2y + z - 4 = 0 is

• 2 : 1

• 4 : 3

• - 1 : 4

• 2 : 3

69.

A plane passes through (2, 3, - 1) and is perpendicular to the line having direction ratios 3, - 4, 7. The perpendicular distance from the origin to this plane is

• $\frac{3}{\sqrt{74}}$

• $\frac{5}{\sqrt{74}}$

• $\frac{6}{\sqrt{74}}$

• $\frac{13}{\sqrt{74}}$

If the angles made by a straight line with thecoordinate axes are, $\mathrm{\alpha }, \frac{\mathrm{\pi }}{2} - \mathrm{\alpha }, \mathrm{\beta }, \mathrm{then} \mathrm{\beta } \mathrm{is} \mathrm{equal} \mathrm{to}$

• 0

• $\frac{\mathrm{\pi }}{6}$

• $\frac{\mathrm{\pi }}{2}$

• $\mathrm{\pi }$