To solve the problem, we need to analyze the given planes and determine the conditions under which they intersect in a line. The planes are defined by the following equations:
1. \( \vec{r} \cdot (\hat{i} + \hat{j} + \hat{k}) = 1 \)
2. \( \vec{r} \cdot (\hat{i} + 2a \hat{j} + \hat{k}) = 2 \)
3. \( \vec{r} \cdot (a \hat{i} + a^2 \hat{j} + \hat{k}) = 3 \)
### Step 1: Identify the normal vectors of the planes
The normal vectors of the planes can be extracted from the coefficients of \( \vec{r} \):
- For the first plane, the normal vector \( \vec{n_1} = \hat{i} + \hat{j} + \hat{k} \) or \( (1, 1, 1) \).
- For the second plane, the normal vector \( \vec{n_2} = \hat{i} + 2a \hat{j} + \hat{k} \) or \( (1, 2a, 1) \).
- For the third plane, the normal vector \( \vec{n_3} = a \hat{i} + a^2 \hat{j} + \hat{k} \) or \( (a, a^2, 1) \).
### Step 2: Condition for intersection in a line
For three planes to intersect in a line, the normal vectors must not be coplanar. This can be checked using the scalar triple product, which can be represented as the determinant of a matrix formed by the normal vectors:
\[
\begin{vmatrix}
1 & 1 & 1 \\
1 & 2a & 1 \\
a & a^2 & 1
\end{vmatrix} = 0
\]
### Step 3: Calculate the determinant
To compute the determinant, we can expand it:
\[
D = 1 \cdot \begin{vmatrix}
2a & 1 \\
a^2 & 1
\end{vmatrix} - 1 \cdot \begin{vmatrix}
1 & 1 \\
a & 1
\end{vmatrix} + 1 \cdot \begin{vmatrix}
1 & 2a \\
a & a^2
\end{vmatrix}
\]
Calculating each of these 2x2 determinants:
1. \( \begin{vmatrix}
2a & 1 \\
a^2 & 1
\end{vmatrix} = 2a \cdot 1 - 1 \cdot a^2 = 2a - a^2 \)
2. \( \begin{vmatrix}
1 & 1 \\
a & 1
\end{vmatrix} = 1 \cdot 1 - 1 \cdot a = 1 - a \)
3. \( \begin{vmatrix}
1 & 2a \\
a & a^2
\end{vmatrix} = 1 \cdot a^2 - 2a \cdot a = a^2 - 2a^2 = -a^2 \)
Putting it all together:
\[
D = 1(2a - a^2) - 1(1 - a) + 1(-a^2) = 2a - a^2 - 1 + a - a^2
\]
Combining like terms:
\[
D = -2a^2 + 3a - 1
\]
### Step 4: Set the determinant to zero
To find the values of \( a \) for which the planes intersect in a line, we set the determinant equal to zero:
\[
-2a^2 + 3a - 1 = 0
\]
### Step 5: Solve the quadratic equation
We can solve this quadratic equation using the quadratic formula \( a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = -2, b = 3, c = -1 \):
\[
D = 3^2 - 4(-2)(-1) = 9 - 8 = 1
\]
Thus,
\[
a = \frac{-3 \pm \sqrt{1}}{2(-2)} = \frac{-3 \pm 1}{-4}
\]
Calculating the two possible values:
1. \( a = \frac{-3 + 1}{-4} = \frac{-2}{-4} = \frac{1}{2} \)
2. \( a = \frac{-3 - 1}{-4} = \frac{-4}{-4} = 1 \)
### Conclusion
The possible values of \( a \) are \( \frac{1}{2} \) and \( 1 \). Therefore, the number of real values of \( a \) is:
**Answer: 2**
