Two vertical poles of heights, 20 m and 80 m stand apart on a horizontal plane. The height (in meters) of the point of intersection of the Lines joining the top of each pole to the foot of the other, from this horizontal plane is :

• 12

• 18

• 15

• 16

The vector equation of the plane through the line of intersection of the planes x + y + z = 1 and 2x + 3y + 4z = 5 which is perpendicular to the plane x – y + z = 0 is :

• $\overrightarrow{\mathrm{r} }. \left(\hat{\mathrm{i}} - \hat{\mathrm{k}}\right) + 2 = 0$

• $\overrightarrow{\mathrm{r}} . \left(\hat{\mathrm{i}} - \hat{\mathrm{k}}\right) - 2 = 0$

• $\overrightarrow{\mathrm{r}} \times \left(\hat{\mathrm{i}} - \hat{\mathrm{k}}\right) + 2 = 0$

The vertices B and C of a $∆$ABC lie on line, $\frac{\mathrm{x} + 2}{3} = \frac{\mathrm{y} - 1}{0} = \frac{\mathrm{z}}{4}$ such that BC = 5 units. Then the area (in sq.units) of this triangle, given that the point A(1, - 1, 2)

• $\sqrt{34}$

• $5\sqrt{17}$

• $2\sqrt{34}$

• 6

If the two lines x + (a - 1)y = 1 and 2x + a²y = 1 $\left(\mathrm{a} \in \mathrm{R} - \left\{0, 1\right\}\right)$ are perpendicular, then the distance of their point of intersection from the origin is

• $\frac{2}{5}$

• $\frac{2}{\sqrt{5}}$

• $\sqrt{\frac{2}{5}}$

• $\frac{\sqrt{2}}{5}$

Let P be the plane, which contains the line of intersection of the planes, x + y + z = 6 and 2x + 3y + z + 5 = 0 and it is perpendicular to the xy - plane. Then the distance of the point (0, 0, 256) from P is equal to

• $63\sqrt{5}$

• $\raisebox{1ex}{$17$}\!\left/ \!\raisebox{-1ex}{$\sqrt{5}$}\right.$

• $205\sqrt{5}$

• $\raisebox{1ex}{$11$}\!\left/ \!\raisebox{-1ex}{$\sqrt{5}$}\right.$

If the line $\frac{\mathrm{x} - 1}{2} = \frac{\mathrm{y} +\hspace{0.17em}1}{3} = \frac{\mathrm{z} - 2}{4}$ meets the plane, x + 2y + 3z = 15 at a point P, then the distance of P from the origin is :

• $\frac{\sqrt{5}}{2}$

• $2\sqrt{5}$

• $\frac{9}{2}$

• $\frac{7}{2}$

A plane passing through the points (0, - 1, 0) and (0, 0, 1) and making an angle $\frac{\mathrm{\pi }}{4}$ with the plane y - z + 5 = 0, also passes through the point :

• $\left(\sqrt{2}, - 1, 4\right)$

• $\left(- \sqrt{2}, - 1, - 4\right)$

• $\left(- \sqrt{2}, 1, - 4\right)$

• $\left(\sqrt{2}, 1, 4\right)$

If the length of perpendicular from point $\left(\mathrm{\beta }, 0, \mathrm{\beta }\right)$, $\left(\mathrm{where} \mathrm{\beta } \ne 0\right)$ to the line $\frac{\mathrm{x}}{1} = \frac{\mathrm{y} - 1}{0} = \frac{\mathrm{z} + 1}{- 1} \mathrm{is} \sqrt{\frac{3}{2}}$, then the value of $\mathrm{\beta }$ is

• - 1

• - 2

• 1

• 2

If Q(0, - 1, - 3) is the image  P in the plane 3x - y + 4z - 2 = 0 and R is the point (3, - 1, - 2). Then the area (in sq. units) of $△$PQR is :

• $\frac{\sqrt{91}}{4}$

• $\frac{\sqrt{91}}{2}$

• $2\sqrt{13}$

• $\frac{\sqrt{65}}{2}$

If the plane 2x - y + 2z = 3 = 0 has the distances $\frac{1}{3} \mathrm{and} \frac{2}{3}$ units from the planes 4x - 2y + 4z + $\mathrm{\lambda }$ = 0  and 2x - y + 2z + $\mathrm{\mu }$, respectively then the maximum value of $\mathrm{\lambda } + \mathrm{\mu }$

• 5

• 15

• 9

• 13