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JEE Main Math Class 12 Three Dimensional Geometry Part 1 Questions
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© examsnet.com
Question : 4
Total: 100
If the point
(
2
,
α
,
β
)
lies on the plane which passes through the points (3,4,2) and (7,0,6) and is perpendicular to the plane
2
x
−
5
y
=
15
,
then
2
α
−
3
β
is equal to :
[Jan. 11, 2019(II)]
12
7
5
17
Validate
Solution:
Let the normal to the required plane is
→
n
,
then
→
n
=
|
^
i
^
j
^
k
4
−
4
4
2
−
5
0
|
=
20
^
i
+
8
^
j
−
12
^
k
∴
Equation of the plane
(
x
−
3
)
×
20
+
(
y
−
4
)
×
8
+
(
z
−
2
)
×
(
−
12
)
=
0
5
x
−
15
+
2
y
−
8
−
3
z
+
6
=
0
5
x
+
2
y
−
3
z
−
17
=
0
.......(1)
Since, equation of plane (1) passes through
(
2
,
α
,
β
)
,
then
10
+
2
α
−
3
β
−
17
=
0
⇒
2
α
−
3
β
=
7
© examsnet.com
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