The radius of the circle given by

x² + y² + z² + 2x - 2y - 2z - 19 = 0

• 4

• 3

• 2

• 1

If x-coordinate of a point P on the line joining the points Q(2, 2, 1) and R(5, 1, - 2) is 4, then the z-coordinate of P is

• - 2

• - 1

• 1

• 2

The equation of the sphere through the points (1, 0, 0), (0, 1, 0) and (1, 1, 1)and having the smallest radius

• $3\left({\mathrm{x}}^2 + {\mathrm{y}}^2 + {\mathrm{z}}^2\right) - 4\mathrm{x} - 4\mathrm{y} - 2\mathrm{z} + 1 = 0$

• $2\left({\mathrm{x}}^2 + {\mathrm{y}}^2 + {\mathrm{z}}^2\right) - 3\mathrm{x} - 3\mathrm{y} - \mathrm{z} + 1 = 0$

• $\left({\mathrm{x}}^2 + {\mathrm{y}}^2 + {\mathrm{z}}^2\right) - \mathrm{x} - \mathrm{y} - \mathrm{z} + 1 = 0$

• ${\mathrm{x}}^2 + {\mathrm{y}}^2 + {\mathrm{z}}^2 - 2\mathrm{x} - 2\mathrm{y} + 4\mathrm{z} + 1 = 0$

74.

The origin is translated to (1, 2). The point(7, 5) in the old system undergoes the following transformations successively.

I. Moves to the new point under the given translation of origin.

II. Translated through 2 units along the negative direction of the new X-axis.

III. Rotated through an angle - about the 4 origin of new system in the clockwise direction. The final position of the point (7, 5) is

• $\left(\frac{9}{\sqrt{2}}, \frac{- 1}{\sqrt{2}}\right)$

• $\left(\frac{7}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)$

• $\left(\frac{7}{\sqrt{2}}, \frac{- 1}{\sqrt{2}}\right)$

• $\left(\frac{5}{\sqrt{2}}, \frac{- 1}{\sqrt{2}}\right)$

The perpendicular distance from the point $\left(1, \mathrm{\pi }\right)$ to the line joining (1, 0⁰) and $\left(1, \frac{\mathrm{\pi }}{2}\right)$ (in polar coordinates) is

• 2

• $\sqrt{3}$

• 1

• $\sqrt{2}$

If D(2, 1, 0), E(2, 0, 0) and F(0, 1, 0) are mid-points of the sides BC, CA and AB of $∆$ABC, respectively. Then, the centroid of$∆$ABC is

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

• $\left(\frac{4}{3}, \frac{2}{3}, 0\right)$

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

• $\left(\frac{2}{3}, \frac{1}{3}, \frac{1}{3}\right)$

If in a $∆$ABC, r₁ = 2, r₂ = 3 and 7a = 6, then ae quals to

• 4

• 1

• 2

• 3

$\mathrm{If} \frac{2{\mathrm{x}}^3 + {\mathrm{x}}^2 - 5}{{\mathrm{x}}^4 - 25} = \frac{\mathrm{Ax} + \mathrm{B}}{{\mathrm{x}}^2 - 5} + \frac{\mathrm{Cx} \hspace{0.17em}+ 1}{{\mathrm{x}}^2 +\hspace{0.17em}5}, \mathrm{then} \left(\mathrm{A}, \mathrm{B}, \mathrm{C}\right) = ?$

• (1, 1, 1)

• (1, 1, 0)

• (1, 0, 1)

• (1, 2, 1)

If in a $∆\mathrm{ABC}$, r₁ = 2r₂ = 3r₃, then the perimeter of the triangle is equal to

• 3a

• 3b

• 3c

• 3(a + b + c)

If the equation to the locus of points equidistant from the points (- 2, 3), (6, - 5) is ax + by + c = 0, where a > 0, then the ascending order of a, b, c is

• a, b, c

• c, b, a

• b, c, a

• a, c, b