1
JEE Main 2021 (Online) 24th February Evening Shift
+4
-1
Let a, b$$\in$$R. If the mirror image of the point P(a, 6, 9) with respect to the line

$${{x - 3} \over 7} = {{y - 2} \over 5} = {{z - 1} \over { - 9}}$$ is (20, b, $$-$$a$$-$$9), then | a + b |, is equal to :
A
88
B
90
C
86
D
84
2
JEE Main 2021 (Online) 24th February Evening Shift
+4
-1
Out of Syllabus
The vector equation of the plane passing through the intersection

of the planes $$\overrightarrow r .\left( {\widehat i + \widehat j + \widehat k} \right) = 1$$ and $$\overrightarrow r .\left( {\widehat i - 2\widehat j} \right) = - 2$$, and the point (1, 0, 2) is :
A
$$\overrightarrow r .\left( {\widehat i + 7\widehat j + 3\widehat k} \right) = {7 \over 3}$$
B
$$\overrightarrow r .\left( {\widehat i + 7\widehat j + 3\widehat k} \right) = 7$$
C
$$\overrightarrow r .\left( {3\widehat i + 7\widehat j + 3\widehat k} \right) = 7$$
D
$$\overrightarrow r .\left( {\widehat i - 7\widehat j + 3\widehat k} \right) = {7 \over 3}$$
3
JEE Main 2021 (Online) 24th February Morning Shift
+4
-1
Out of Syllabus
The equation of the plane passing through the point (1, 2, -3) and perpendicular to the planes

3x + y - 2z = 5 and 2x - 5y - z = 7, is :
A
6x - 5y + 2z + 10 =0
B
3x - 10y - 2z + 11 = 0
C
6x - 5y - 2z - 2 = 0
D
11x + y + 17z + 38 = 0
4
JEE Main 2021 (Online) 24th February Morning Shift
+4
-1
Out of Syllabus
The distance of the point (1, 1, 9) from the point of intersection of the line $${{x - 3} \over 1} = {{y - 4} \over 2} = {{z - 5} \over 2}$$ and the plane x + y + z = 17 is :
A
$$19\sqrt 2$$
B
$$2\sqrt {19}$$
C
38
D
$$\sqrt {38}$$
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