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Test Index
JEE Main 1-Feb-2024 Shift 2 Solved Paper
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Section:
Mathematics
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© examsnet.com
Question : 8
Total: 90
Let
P
and
Q
be the points on the line
x
+
3
8
=
y
−
4
2
=
z
+
1
2
which are at a distance of 6 units from the point
R
(
1
,
2
,
3
)
. If the centroid of the triangle PQR is
(
α
,
β
,
γ
)
, then
α
2
+
β
2
+
γ
2
is:
[1-Feb-2024 Shift 2]
26
36
18
24
Validate
Solution:
👈: Video Solution
P
(
8
λ
−
3
,
2
λ
+
4
,
2
λ
−
1
)
PR
=
6
(
8
λ
−
4
)
2
+
(
2
λ
+
2
)
2
+
(
2
λ
−
4
)
2
=
36
λ
=
0
,
1
Hence
P
(
−
3
,
4
,
−
1
)
&
Q
(
5
,
6
,
1
)
Centroid of
∆
PQR
=
(
1
,
4
,
1
)
≡
(
α
,
β
,
γ
)
α
2
+
β
2
+
γ
2
=
18
© examsnet.com
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