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

If I is incentre of $∆$ABC, then I is

• $\frac{\mathrm{a}\overrightarrow{\mathrm{a}} +\hspace{0.17em}\mathrm{b}\overrightarrow{\mathrm{b}} +\hspace{0.17em}\mathrm{c}\overrightarrow{\mathrm{c}}}{\mathrm{a} +\hspace{0.17em}\mathrm{b} +\hspace{0.17em}\mathrm{c}}$

• $\frac{\mathrm{a}\overrightarrow{\mathrm{a}} +\hspace{0.17em}\mathrm{b}\overrightarrow{\mathrm{b}} +\hspace{0.17em}\mathrm{c}\overrightarrow{\mathrm{c}}}{\sqrt{{\mathrm{a}}^2 +\hspace{0.17em}{\mathrm{b}}^2 +\hspace{0.17em}{\mathrm{c}}^2}}$

• $\frac{1}{3}\left[\overrightarrow{\mathrm{a}} + \overrightarrow{\mathrm{b}} + \overrightarrow{\mathrm{c}}\right]$

• $\frac{\overrightarrow{\mathrm{a}} +\hspace{0.17em}\overrightarrow{\mathrm{b}} + \overrightarrow{\mathrm{c}}}{\mathrm{a} +\hspace{0.17em}\mathrm{b} +\hspace{0.17em}\mathrm{c}}$

What are the DR's of vector parallel to (2, - 1, 1) and (3 4, - 1) ?

• (1, 5, - 2)

• (- 2, - 5, 2)

• (- 1, 5, 2)

• (- 1, - 5, - 2)

Let ABCD be a parallelogram whose diagonals intersect at P and O be the origin, then $\overrightarrow{\mathrm{OA}} + \overrightarrow{\mathrm{OB}} + \overrightarrow{\mathrm{OC}} + \overrightarrow{\mathrm{OD}}$

• $\overrightarrow{\mathrm{OP}}$

• $2\overrightarrow{\mathrm{OP}}$

• $3\overrightarrow{\mathrm{OP}}$

• $4\overrightarrow{\mathrm{OP}}$

A unit vector in the plane of $\hat{\mathrm{i}} +\hspace{0.17em}2\hat{\mathrm{j}} +\hspace{0.17em}\hat{\mathrm{k}}$ and $\hat{\mathrm{i}} +\hspace{0.17em}\hat{\mathrm{j}} +\hspace{0.17em}2\hat{\mathrm{k}}$ and perpendicular to $2\hat{\mathrm{i}} +\hspace{0.17em}\hat{\mathrm{j}} +\hspace{0.17em}\hat{\mathrm{k}}$, is

• $\hat{\mathrm{j}} -\hspace{0.17em}\hat{\mathrm{k}}$

• $\frac{\hat{\mathrm{i}} +\hspace{0.17em}\hat{\mathrm{j}}}{\sqrt{2}}$

• $\frac{\hat{\mathrm{j}} +\hspace{0.17em}\hat{\mathrm{k}}}{\sqrt{2}}$

• $\frac{\hat{\mathrm{j}} -\hspace{0.17em}\hat{\mathrm{k}}}{\sqrt{2}}$

569.

The line joining the points $6\overrightarrow{\mathrm{a}} - 4\overrightarrow{\mathrm{b}} + 4\overrightarrow{\mathrm{c}}, - 4\overrightarrow{\mathrm{c}}$ and the line joining the points $- \overrightarrow{\mathrm{a}} - 2\overrightarrow{\mathrm{b}} - 3\overrightarrow{\mathrm{c}}, \overrightarrow{\mathrm{a}} + 2\overrightarrow{\mathrm{b}} - 5\overrightarrow{\mathrm{c}}$ intersect at

• $- 4\overrightarrow{\mathrm{a}}$

• $4\overrightarrow{\mathrm{a}} - \overrightarrow{\mathrm{b}} - \overrightarrow{\mathrm{c}}$

• $4\overrightarrow{\mathrm{c}}$

• None of the above

The symmetric equation of lines 3x + 2y + z - 5 = 0 and x + y - 2z - 3 = 0, is

• $\frac{\mathrm{x} - 1}{5} = \frac{\mathrm{y} - 4}{7} = \frac{\mathrm{z} - 0}{1}$

• $\frac{\mathrm{x} + 1}{5} = \frac{\mathrm{y} + 4}{7} = \frac{\mathrm{z} - 0}{1}$

• $\frac{\mathrm{x} + 1}{- 5} = \frac{\mathrm{y} - 4}{7} = \frac{\mathrm{z} - 0}{1}$

• $\frac{\mathrm{x} - 1}{- 5} = \frac{\mathrm{y} - 4}{7} = \frac{\mathrm{z} - 0}{1}$