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Question
Mathematics
Let the equation of the plane passing through the line x-2 y-z-5=0=x+y+3 z-5 and parallel to the line x+y+2 z-7=0=2 x+3 y+z-2 be a x+b y+c z=65. Then the distance of the point (a, b, c) from the plane 2 x+2 y-z+16=0 is
Q. Let the equation of the plane passing through the line
x
−
2
y
−
z
−
5
=
0
=
x
+
y
+
3
z
−
5
and parallel to the line
x
+
y
+
2
z
−
7
=
0
=
2
x
+
3
y
+
z
−
2
be
a
x
+
b
y
+
cz
=
65
. Then the distance of the point
(
a
,
b
,
c
)
from the plane
2
x
+
2
y
−
z
+
16
=
0
is _______
659
141
JEE Main
JEE Main 2023
Three Dimensional Geometry
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Answer:
9
Solution:
Equation of plane is
(
x
−
2
y
−
z
−
5
)
+
b
(
x
+
y
+
3
z
−
5
)
=
0
∣
∣
1
+
b
1
2
−
2
+
b
1
3
−
1
+
3
b
2
1
∣
∣
=
0
⇒
b
=
12
∴
plane is
13
x
+
10
y
+
35
z
=
65
Distance from given point to plane
=
9