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JEE Mains 1-Feb-2023 Shift 2 Solved Paper
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© examsnet.com
Question : 69
Total: 90
Let the plane P pass through the intersection of the planes
2
x
+
3
y
−
z
=
2
and
x
+
2
y
+
3
z
=
6
, and be perpendicular to the plane
2
x
+
y
−
z
+
1
=
0
. If
d
is the distance of
P
from the point
(
−
7
,
1
,
1
)
, then
d
2
is equal to :
[1-Feb-2023 Shift 2]
250
83
15
53
25
83
250
82
Validate
Solution:
P
≡
P
1
+
λ
P
2
=
0
(
2
+
λ
)
x
+
(
3
+
2
λ
)
y
+
(
3
λ
−
1
)
z
−
2
−
6
λ
=
0
Plane
P
is perpendicular to
P
3
∴
→
n
⋅
→
n
3
=
0
2
(
λ
+
2
)
+
(
2
λ
+
3
)
−
(
3
λ
−
1
)
=
0
λ
=
−
8
P
≡
−
6
x
−
13
y
−
25
z
+
46
=
0
6
x
+
13
y
+
25
z
−
46
=
0
Dist from
(
−
7
,
1
,
1
)
d
=
∣
−
42
+
13
+
25
−
46
√
36
+
169
+
625
∣
=
50
√
830
d
2
=
50
×
50
830
=
250
83
© examsnet.com
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