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Mathematics
The equation of the plane passing through the point (1, 2, -3) and perpendicular to the planes 3 x+y-2 z=5 and 2 x-5 y-z=7 is
Q. The equation of the plane passing through the point (1, 2, -3) and perpendicular to the planes
3
x
+
y
−
2
z
=
5
and
2
x
−
5
y
−
z
=
7
is
1795
234
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JEE Main 2021
Three Dimensional Geometry
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A
3
x
−
10
y
−
2
z
+
11
=
0
50%
B
6
x
−
5
y
−
2
z
−
2
=
0
0%
C
11
x
+
y
+
17
z
+
38
=
0
50%
D
6
x
−
5
y
+
2
z
+
10
=
0
0%
Solution:
Normal vector :
∣
∣
i
^
3
2
j
^
1
−
5
k
^
−
2
−
1
∣
∣
=
−
11
i
^
−
j
^
+
17
k
^
So drs of normal to the required plane is
<
11
,
1
,
17
>
plane passes through (1,2,-3) So eq
n
of plane:
11
(
x
−
1
)
+
1
(
y
−
2
)
+
17
(
z
+
3
)
=
0
⇒
11
x
+
y
+
17
z
+
38
=
0