1
JEE Main 2017 (Online) 8th April Morning Slot
+4
-1
Out of Syllabus
The coordinates of the foot of the perpendicular from the point (1, $$-$$2, 1) on the plane containing the lines, $${{x + 1} \over 6} = {{y - 1} \over 7} = {{z - 3} \over 8}$$ and $${{x - 1} \over 3} = {{y - 2} \over 5} = {{z - 3} \over 7},$$ is :
A
(2, $$-$$4, 2)
B
($$-$$ 1, 2, $$-$$1)
C
(0, 0, 0)
D
(1, 1, 1)
2
JEE Main 2017 (Online) 8th April Morning Slot
+4
-1
Out of Syllabus
The line of intersection of the planes $$\overrightarrow r .\left( {3\widehat i - \widehat j + \widehat k} \right) = 1\,\,$$ and
$$\overrightarrow r .\left( {\widehat i + 4\widehat j - 2\widehat k} \right) = 2,$$ is :
A
$${{x - {4 \over 7}} \over { - 2}} = {y \over 7} = {{z - {5 \over 7}} \over {13}}$$
B
$${{x - {4 \over 7}} \over 2} = {y \over { - 7}} = {{z + {5 \over 7}} \over {13}}$$
C
$${{x - {6 \over {13}}} \over 2} = {{y - {5 \over {13}}} \over { - 7}} = {z \over { - 13}}$$
D
$${{x - {6 \over {13}}} \over 2} = {{y - {5 \over {13}}} \over 7} = {z \over { - 13}}$$
3
JEE Main 2017 (Offline)
+4
-1
Out of Syllabus
If the image of the point P(1, –2, 3) in the plane, 2x + 3y – 4z + 22 = 0 measured parallel to the line,

$${x \over 1} = {y \over 4} = {z \over 5}$$ is Q, then PQ is equal to:
A
$$2\sqrt {42}$$
B
$$\sqrt {42}$$
C
$$6\sqrt 5$$
D
$$3\sqrt 5$$
4
JEE Main 2017 (Offline)
+4
-1
Out of Syllabus
The distance of the point (1, 3, – 7) from the plane passing through the point (1, –1, – 1), having normal perpendicular to both the lines

$${{x - 1} \over 1} = {{y + 2} \over { - 2}} = {{z - 4} \over 3}$$

and

$${{x - 2} \over 2} = {{y + 1} \over { - 1}} = {{z + 7} \over { - 1}}$$ is :
A
$${{10} \over {\sqrt {83} }}$$
B
$${{5} \over {\sqrt {83} }}$$
C
$${{10} \over {\sqrt {74} }}$$
D
$${{20} \over {\sqrt {74} }}$$
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