The values of α, for which $\left|\begin{array}{ccc}1& \frac{3}{2}& \mathrm{\alpha <!--\; \alpha \; -->}+\frac{3}{2}\\ 1& \frac{1}{3}& \mathrm{\alpha <!--\; \alpha \; -->}+\frac{1}{3}\\ 2\mathrm{\alpha <!--\; \alpha \; -->}+3& 3\mathrm{\alpha <!--\; \alpha \; -->}+1& 0\end{array}\right|=0$ lie in the interval

The correct answer is (–3, 0).

We need to find the values of α for which the following determinant equals zero:

D=\left|\begin{array}{ccc}1& \frac{3}{2}& \alpha +\frac{3}{2}\\ 1& \frac{1}{3}& \mathrm{��}+\frac{1}{3}\\ 2\alpha +3& 3\alpha +1& 0\end{array}\right|=0

Step 1: Apply Row Operation R₁ → R₁ - R₂

Subtracting Row 2 from Row 1 element-by-element:

- Element (1,1): 1-1=0

- Element (1,2): \frac{3}{2}-\frac{1}{3}=\frac{9}{6}-\frac{2}{6}=\frac{7}{6}

- Element (1,3): (\alpha +\frac{3}{2})-(\alpha +\frac{1}{3})=\frac{3}{2}-\frac{1}{3}=\frac{7}{6}

So the matrix becomes:

\left|\begin{array}{ccc}0& \frac{7}{6}& \frac{7}{6}\\ 1& \frac{1}{3}& \alpha +\frac{1}{3}\\ 2\alpha +3& 3\alpha +1& 0\end{array}\right|

Step 2: Expand Along the First Column

Expanding along Column 1 (only the 2nd and 3rd row elements are non-zero in column 1):

D=-1·\left|\begin{array}{cc}\frac{7}{6}& \frac{7}{6}\\ 3\alpha +1& 0\end{array}\right|+(2\alpha +3)·\left|\begin{array}{cc}\frac{7}{6}& \frac{7}{6}\\ \frac{1}{3}& \alpha +\frac{1}{3}\end{array}\right|

Step 3: Evaluate Each 2×2 Determinant

First 2×2 determinant:

\left|\begin{array}{cc}\frac{7}{6}& \frac{7}{6}\\ 3\alpha +1& 0\end{array}\right|=\frac{7}{6}·0-\frac{7}{6}·(3\alpha +1)=-\frac{7}{6}(3\alpha +1)

Second 2×2 determinant:

\left|\begin{array}{cc}\frac{7}{6}& \frac{7}{6}\\ \frac{1}{3}& \alpha +\frac{1}{3}\end{array}\right|=\frac{7}{6}(\alpha +\frac{1}{3})-\frac{7}{6}·\frac{1}{3}=\frac{7}{6}(\alpha +\frac{1}{3}-\frac{1}{3})=\frac{7}{6}\alpha

Step 4: Substitute Back

D=-1·[-\frac{7}{6}(3\alpha +1\left)\right]+(2\alpha +3)·\frac{7}{6}\alpha

=\frac{7}{6}(3\alpha +1)+\frac{7}{6}\alpha (2\alpha +3)

Factor out \frac{7}{6}:

D=\frac{7}{6}\left[\right(3\alpha +1)+\alpha (2\alpha +3\left)\right]

Expand inside the brackets:

=\frac{7}{6}[3\alpha +1+2{\alpha }^2+3\alpha ]=\frac{7}{6}[2{\alpha }^2+6\alpha +1]

Step 5: Set D = 0 and Solve for α

Since \frac{7}{6}\ne \; 0, we need:

2{\alpha }^2+6\alpha +1=0

Using the quadratic formula:

\alpha =\frac{-6\pm \sqrt{36-8}}{2·2}=\frac{-6\pm \sqrt{28}}{4}=\frac{-6\pm 2\sqrt{7}}{4}=\frac{-3\pm \sqrt{7}}{2}

Step 6: Compute the Two Roots

Since \sqrt{7}\approx 2.646:

{\alpha }_1=\frac{-3+\sqrt{7}}{2}\approx \frac{-3+2.646}{2}\approx \frac{-0.354}{2}\approx -0.177

{\alpha }_2=\frac{-3-\sqrt{7}}{2}\approx \frac{-3-2.646}{2}\approx \frac{-5.646}{2}\approx -2.823

Step 7: Identify the Interval

Both roots are:

{\alpha }_1\approx -0.177\text{and}{\alpha }_2\approx -2.823

Checking each against the given options:

- Both -0.177 and -2.823 lie within the interval (−3, 0).

- Neither value lies in (-\frac{3}{2},\frac{3}{2}) since -2.823 is outside that range.

- Neither value is positive, so (0, 3) is ruled out.

- For (−2, 1): -2.823 is outside this range.

Therefore, the interval that contains both values of α is (−3, 0).