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Byju's Answer
Standard XII
Mathematics
Equation of Circle in Complex Form
Evaluate | ...
Question
Evaluate
∣
∣ ∣
∣
x
C
1
x
C
2
x
C
3
y
C
1
y
C
2
y
C
3
z
C
1
z
C
2
z
C
3
∣
∣ ∣
∣
.
Open in App
Solution
Δ
=
∣
∣ ∣ ∣
∣
X
X
(
X
−
1
)
/
1.2
X
(
X
−
1
)
(
X
−
2
)
/
1.2.3
Y
Y
(
Y
−
1
)
/
1.2
Y
(
Y
−
1
)
(
Y
−
2
)
/
1.2.3
Z
Z
(
Z
−
1
)
/
1.2
Z
(
Z
−
1
)
(
Z
−
2
)
/
1.2.3
∣
∣ ∣ ∣
∣
=
X
Y
Z
2.6
∣
∣ ∣ ∣
∣
1
(
X
−
1
)
(
X
−
1
)
(
X
−
2
)
1
(
Y
−
1
)
(
Y
−
1
)
(
Y
−
2
)
1
(
Z
−
1
)
(
Z
−
1
)
(
Z
−
2
)
∣
∣ ∣ ∣
∣
[Operate
R
2
−
R
1
and
R
3
−
R
1
and expend with
C
1
]
=
X
Y
Z
12
∣
∣ ∣ ∣
∣
1
(
X
−
1
)
(
X
−
1
)
(
X
−
2
)
0
(
Y
−
X
)
(
Y
−
X
)
(
Y
+
X
−
3
)
0
(
Z
−
X
)
(
Z
−
X
)
(
Z
+
X
−
3
)
∣
∣ ∣ ∣
∣
=
X
Y
Z
12
.
(
Y
−
X
)
(
Z
−
X
)
∣
∣
∣
1
Y
+
X
−
3
1
Z
+
X
−
3
∣
∣
∣
=
X
Y
Z
12
(
Y
−
X
)
(
Z
−
X
)
[
(
Z
+
X
−
3
)
−
(
Y
+
X
−
3
)
]
=
X
Y
Z
12
(
Y
−
X
)
(
Z
−
X
)
(
Z
−
Y
)
=
1
12
X
Y
Z
(
Y
−
Z
)
(
Z
−
X
)
(
X
−
Y
)
.
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0
Similar questions
Q.
The determinant
∣
∣ ∣
∣
x
C
1
x
C
2
x
C
3
y
C
1
y
C
2
y
C
3
z
C
1
z
C
2
z
C
3
∣
∣ ∣
∣
=
Q.
Eliminate x, y, z from the equations
a
1
x
+
b
1
y
+
c
1
z
=
0
.
.
.
(
1
)
,
a
2
x
+
b
2
y
+
c
2
z
=
0
.
.
.
(
2
)
,
a
3
x
+
b
3
y
+
c
3
z
=
0
.
.
.
(
3
)
Q.
If
A
=
∣
∣ ∣
∣
a
1
b
1
c
1
a
2
b
2
c
2
a
3
b
3
a
3
∣
∣ ∣
∣
≠
0
,
then the system of equations
a
1
x
+
b
1
y
+
c
1
z
=
0
,
a
2
x
+
b
2
y
+
c
2
z
=
0
and
a
3
x
+
b
3
y
+
c
3
z
=
0
has:
Q.
Consider the system of equations
a
1
x
+
b
1
y
+
c
1
z
=
0
a
2
x
+
b
2
y
+
c
2
z
=
0
a
3
x
+
b
3
y
+
c
3
z
=
0
i
f
∣
∣ ∣
∣
a
1
b
1
c
1
a
2
b
2
c
3
a
3
b
3
c
3
∣
∣ ∣
∣
=
0
,
then the system has
Q.
If
f
(
x
)
=
∣
∣ ∣
∣
x
+
c
1
x
+
a
x
+
a
x
+
b
x
+
c
2
x
+
a
x
+
b
x
+
b
x
+
c
3
∣
∣ ∣
∣
, then
f
(
x
)
is
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