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JEE Main Math Class 12 Three Dimensional Geometry Part 2 Questions
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© examsnet.com
Question : 5
Total: 100
Let
x
−
2
3
=
y
+
1
−
2
=
z
+
3
−
1
lie on the plane
p
x
−
q
y
+
z
=
5
, for some
p
,
q
∈
R
. The shortest distance of the plane from the origin is :
[29-Jun-2022-Shift-2]
√
3
109
√
5
142
5
√
71
1
√
142
Validate
Solution:
(
2
,
−
1
,
−
3
)
satisfy the given plane.
So
2
p
+
q
=
8
Also given line is perpendicular to normal plane so
3
p
+
2
q
−
1
=
0
⇒
p
=
15
,
q
=
−
22
Eq. of plane
15
x
−
22
y
+
z
−
5
=
0
its distance from origin
=
6
√
710
=
√
5
142
© examsnet.com
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