Let
D = ∣ 1 3 2 α + 3 2 1 1 3 α + 1 3 2 α + 3 3 α + 1 0 ∣ . D=\begin{vmatrix}
1 & \frac{3}{2} & \alpha+\frac{3}{2}\\
1 & \frac{1}{3} & \alpha+\frac{1}{3}\\
2\alpha+3 & 3\alpha+1 & 0
\end{vmatrix}. D = 1 1 2 α + 3 2 3 3 1 3 α + 1 α + 2 3 α + 3 1 0 . We need the values of α \alpha α D = 0 D=0 D = 0
Apply the row operation
R 1 → R 1 − R 2 . R_1 \to R_1-R_2. R 1 → R 1 − R 2 .
D = ∣ 0 3 2 − 1 3 ( α + 3 2 ) − ( α + 1 3 ) 1 1 3 α + 1 3 2 α + 3 3 α + 1 0 ∣ . D=\begin{vmatrix}
0 & \frac{3}{2}-\frac{1}{3} & \left(\alpha+\frac{3}{2}\right)-\left(\alpha+\frac{1}{3}\right)\\
1 & \frac{1}{3} & \alpha+\frac{1}{3}\\
2\alpha+3 & 3\alpha+1 & 0
\end{vmatrix}. D = 0 1 2 α + 3 2 3 − 3 1 3 1 3 α + 1 ( α + 2 3 ) − ( α + 3 1 ) α + 3 1 0 . Now,
3 2 − 1 3 = 9 − 2 6 = 7 6 , ( α + 3 2 ) − ( α + 1 3 ) = 7 6 . \frac{3}{2}-\frac{1}{3}=\frac{9-2}{6}=\frac{7}{6},
\qquad
\left(\alpha+\frac{3}{2}\right)-\left(\alpha+\frac{1}{3}\right)=\frac{7}{6}. 2 3 − 3 1 = 6 9 − 2 = 6 7 , ( α + 2 3 ) − ( α + 3 1 ) = 6 7 . So
D = ∣ 0 7 6 7 6 1 1 3 α + 1 3 2 α + 3 3 α + 1 0 ∣ . D=\begin{vmatrix}
0 & \frac{7}{6} & \frac{7}{6}\\
1 & \frac{1}{3} & \alpha+\frac{1}{3}\\
2\alpha+3 & 3\alpha+1 & 0
\end{vmatrix}. D = 0 1 2 α + 3 6 7 3 1 3 α + 1 6 7 α + 3 1 0 .
Factor out 7 6 \frac{7}{6} 6 7
D = 7 6 ∣ 0 1 1 1 1 3 α + 1 3 2 α + 3 3 α + 1 0 ∣ . D=\frac{7}{6}
\begin{vmatrix}
0 & 1 & 1\\
1 & \frac{1}{3} & \alpha+\frac{1}{3}\\
2\alpha+3 & 3\alpha+1 & 0
\end{vmatrix}. D = 6 7 0 1 2 α + 3 1 3 1 3 α + 1 1 α + 3 1 0 . Thus D = 0 D=0 D = 0
∣ 0 1 1 1 1 3 α + 1 3 2 α + 3 3 α + 1 0 ∣ = 0. \begin{vmatrix}
0 & 1 & 1\\
1 & \frac{1}{3} & \alpha+\frac{1}{3}\\
2\alpha+3 & 3\alpha+1 & 0
\end{vmatrix}=0. 0 1 2 α + 3 1 3 1 3 α + 1 1 α + 3 1 0 = 0.
Expand along the first row:
Δ = 0 ⋅ C 11 + 1 ⋅ C 12 + 1 ⋅ C 13 . \Delta=
0\cdot C_{11}+1\cdot C_{12}+1\cdot C_{13}. Δ = 0 ⋅ C 11 + 1 ⋅ C 12 + 1 ⋅ C 13 . So
Δ = − ∣ 1 α + 1 3 2 α + 3 0 ∣ + ∣ 1 1 3 2 α + 3 3 α + 1 ∣ . \Delta=-\begin{vmatrix}1 & \alpha+\frac{1}{3}\\ 2\alpha+3 & 0\end{vmatrix}
+\begin{vmatrix}1 & \frac{1}{3}\\ 2\alpha+3 & 3\alpha+1\end{vmatrix}. Δ = − 1 2 α + 3 α + 3 1 0 + 1 2 α + 3 3 1 3 α + 1 . Now compute each minor:
∣ 1 α + 1 3 2 α + 3 0 ∣ = 1 ⋅ 0 − ( α + 1 3 ) ( 2 α + 3 ) = − ( α + 1 3 ) ( 2 α + 3 ) . \begin{vmatrix}1 & \alpha+\frac{1}{3}\\ 2\alpha+3 & 0\end{vmatrix}
=1\cdot 0-(\alpha+\tfrac13)(2\alpha+3)
=-(\alpha+\tfrac13)(2\alpha+3). 1 2 α + 3 α + 3 1 0 = 1 ⋅ 0 − ( α + 3 1 ) ( 2 α + 3 ) = − ( α + 3 1 ) ( 2 α + 3 ) . Hence
− ∣ 1 α + 1 3 2 α + 3 0 ∣ = ( α + 1 3 ) ( 2 α + 3 ) . -\begin{vmatrix}1 & \alpha+\frac{1}{3}\\ 2\alpha+3 & 0\end{vmatrix}
=(\alpha+\tfrac13)(2\alpha+3). − 1 2 α + 3 α + 3 1 0 = ( α + 3 1 ) ( 2 α + 3 ) . Also,
∣ 1 1 3 2 α + 3 3 α + 1 ∣ = 1 ( 3 α + 1 ) − 1 3 ( 2 α + 3 ) = 3 α + 1 − 2 α + 3 3 . \begin{vmatrix}1 & \frac{1}{3}\\ 2\alpha+3 & 3\alpha+1\end{vmatrix}
=1(3\alpha+1)-\frac13(2\alpha+3)
=3\alpha+1-\frac{2\alpha+3}{3}. 1 2 α + 3 3 1 3 α + 1 = 1 ( 3 α + 1 ) − 3 1 ( 2 α + 3 ) = 3 α + 1 − 3 2 α + 3 . Simplify:
3 α + 1 − 2 α + 3 3 = 9 α + 3 − 2 α − 3 3 = 7 α 3 . 3\alpha+1-\frac{2\alpha+3}{3}
=\frac{9\alpha+3-2\alpha-3}{3}
=\frac{7\alpha}{3}. 3 α + 1 − 3 2 α + 3 = 3 9 α + 3 − 2 α − 3 = 3 7 α . Therefore,
Δ = ( α + 1 3 ) ( 2 α + 3 ) + 7 α 3 . \Delta=(\alpha+\tfrac13)(2\alpha+3)+\frac{7\alpha}{3}. Δ = ( α + 3 1 ) ( 2 α + 3 ) + 3 7 α .
Expand:
( α + 1 3 ) ( 2 α + 3 ) = 2 α 2 + 3 α + 2 α 3 + 1 = 2 α 2 + 11 α 3 + 1. (\alpha+\tfrac13)(2\alpha+3)=2\alpha^2+3\alpha+\frac{2\alpha}{3}+1
=2\alpha^2+\frac{11\alpha}{3}+1. ( α + 3 1 ) ( 2 α + 3 ) = 2 α 2 + 3 α + 3 2 α + 1 = 2 α 2 + 3 11 α + 1. So
Δ = 2 α 2 + 11 α 3 + 1 + 7 α 3 = 2 α 2 + 6 α + 1. \Delta=2\alpha^2+\frac{11\alpha}{3}+1+\frac{7\alpha}{3}
=2\alpha^2+6\alpha+1. Δ = 2 α 2 + 3 11 α + 1 + 3 7 α = 2 α 2 + 6 α + 1. Hence
D = 0 ⟺ 2 α 2 + 6 α + 1 = 0. D=0 \iff 2\alpha^2+6\alpha+1=0. D = 0 ⟺ 2 α 2 + 6 α + 1 = 0.
Solve the quadratic:
2 α 2 + 6 α + 1 = 0 2\alpha^2+6\alpha+1=0 2 α 2 + 6 α + 1 = 0 α = − 6 ± 36 − 8 4 = − 6 ± 28 4 = − 6 ± 2 7 4 = − 3 ± 7 2 . \alpha=\frac{-6\pm\sqrt{36-8}}{4}
=\frac{-6\pm\sqrt{28}}{4}
=\frac{-6\pm 2\sqrt7}{4}
=\frac{-3\pm\sqrt7}{2}. α = 4 − 6 ± 36 − 8 = 4 − 6 ± 28 = 4 − 6 ± 2 7 = 2 − 3 ± 7 .
Approximate the roots:
7 ≈ 2.646 , \sqrt7\approx 2.646, 7 ≈ 2.646 , so
α 1 = − 3 − 7 2 ≈ − 5.646 2 ≈ − 2.823 , \alpha_1=\frac{-3-\sqrt7}{2}\approx \frac{-5.646}{2}\approx -2.823, α 1 = 2 − 3 − 7 ≈ 2 − 5.646 ≈ − 2.823 , α 2 = − 3 + 7 2 ≈ − 0.354 2 ≈ − 0.177. \alpha_2=\frac{-3+\sqrt7}{2}\approx \frac{-0.354}{2}\approx -0.177. α 2 = 2 − 3 + 7 ≈ 2 − 0.354 ≈ − 0.177. Thus the values of α \alpha α − 2.823 -2.823 − 2.823 − 0.177 -0.177 − 0.177
Now check the options for the set of these values:
A: ( − 2 , 1 ) (-2,1) ( − 2 , 1 ) − 0.177 -0.177 − 0.177 − 2.823 -2.823 − 2.823
B: ( − 3 2 , 3 2 ) \left(-\frac32,\frac32\right) ( − 2 3 , 2 3 ) − 0.177 -0.177 − 0.177 − 2.823 -2.823 − 2.823
C: ( − 3 , 0 ) (-3,0) ( − 3 , 0 ) − 2.823 -2.823 − 2.823 − 0.177 -0.177 − 0.177
D: ( 0 , 3 ) (0,3) ( 0 , 3 )
Therefore, the correct option is
C ( − 3 , 0 ) . \boxed{\text{C }(-3,0)}. C ( − 3 , 0 ) .