The line joining the points `A(2,1)`, and `B(3,2)` is perpendicular to the line `(a^2)x+(a+2)y+2=0`. Find the values of a.

Explanation

To solve the problem, we need to find the values of \( a \) such that the line joining the points \( A(2,1) \) and \( B(3,2) \) is perpendicular to the line given by the equation \( (a^2)x + (a + 2)y + 2 = 0 \).

### Step 1: Find the slope of the line joining points A and B
The slope \( m_{AB} \) of the line joining the points \( A(2,1) \) and \( B(3,2) \) is calculated using the formula:
\[ m_{AB} = \frac{y_2 - y_1}{x_2 - x_1} \]
Substituting the coordinates of points A and B:
\[ m_{AB} = \frac{2 - 1}{3 - 2} = \frac{1}{1} = 1 \]

### Step 2: Find the slope of the line given by the equation
The given line is in the form \( ax + by + c = 0 \). To find its slope, we can rearrange it into the slope-intercept form \( y = mx + b \).
The equation is:
\[ (a^2)x + (a + 2)y + 2 = 0 \]
Rearranging gives:
\[ (a + 2)y = -a^2x - 2 \]
\[ y = -\frac{a^2}{a + 2}x - \frac{2}{a + 2} \]
Thus, the slope \( m_L \) of the line is:
\[ m_L = -\frac{a^2}{a + 2} \]

### Step 3: Set up the condition for perpendicular lines
Since the lines are perpendicular, the product of their slopes must equal \(-1\):
\[ m_{AB} \cdot m_L = -1 \]
Substituting the slopes we found:
\[ 1 \cdot \left(-\frac{a^2}{a + 2}\right) = -1 \]
This simplifies to:
\[ -\frac{a^2}{a + 2} = -1 \]
Removing the negative sign:
\[ \frac{a^2}{a + 2} = 1 \]

### Step 4: Solve for a
Cross-multiplying gives:
\[ a^2 = a + 2 \]
Rearranging this equation:
\[ a^2 - a - 2 = 0 \]
Now we can factor this quadratic equation:
\[ (a - 2)(a + 1) = 0 \]
Setting each factor to zero gives the solutions:
\[ a - 2 = 0 \quad \Rightarrow \quad a = 2 \]
\[ a + 1 = 0 \quad \Rightarrow \quad a = -1 \]

### Final Answer
The values of \( a \) are:
\[ \boxed{2} \quad \text{and} \quad \boxed{-1} \]