Question - Determining the Constant Value for a System of Equations Without Solutions

The task is to determine the value of \( p \) if the system has no solution. A system of equations has no solution when the coefficients of the variables can be made proportional, while the constants are not proportional (this creates a situation where the lines or planes are parallel, but not coincident).

First, we should identify if there's a set of multiples that can make the first two rows look similar. If we multiply the first row by 2, the x and y coefficients will match the second row:

This matches the second equation \( \frac{2}{3}x + \frac{4}{3}y - z = 0 \). Let's write the two equations side by side:

To determine when the system has no solutions, we need to make the coefficients on the left sides proportional but ensure the constants are not. From the transformed first row and the second row, we can see that to make the \(x\) terms proportional, I can multiply the second equation by 9:

Since \( 9x = 6x \), the \(x\) coefficients are now proportional. We can do the same for the second row and verify it is true for the \(y\) terms:

Comparing the new \(y\) term \(12y\) with the first equation, we notice that the coefficient of \(y\) in the first equation \( \frac{2}{3}y \) is now proportional, and so is the term involving \(z\) since we added a \(-z\) term in the second equation to match with the first equation's \( \frac{4}{3}z \), making them proportional as well. Since we already ensured that the constants are not proportional (the constant in the second row is 0), the system should have no solution. However, we need to check the third equation to ensure this holds true for all three equations.

For the system to have no solution, the third equation must also follow the pattern of proportional coefficients with a not-proportional constant. For that, I need to choose a suitable value for \( p \) that will make the coefficients of \(x\) and \(y\) in the third equation proportional to the corresponding coefficients in the first two equations, but the constant term \( \frac{3}{2} \) should not be proportional to \( 0 \) and \( \frac{4}{3} \) (otherwise, the system would be consistent).

Since we want the \(x\) terms in the third equation to have a coefficient of \(3\) when the other two are 6 and \(9x\) (which are already proportional), the third equation's \(x\) coefficient is already proportional. For the \(y\) term, \( \frac{2}{3}y \) in the third equation when compared to \( \frac{4}{3}y \) in the first equation and \(12y\) in the second equation is already proportional as well. What remains is the coefficient for \(z\); since there is a \(-z\) term and a \( \frac{4}{3}z \) term in the previously analyzed equations, the \(z\) term in the third equation should also relate to \( -1 \) and \( \frac{4}{3} \).

This means we should find \( p \) such that \( pz \) is proportional to \( -z \) and \( \frac{4}{3}z \). Therefore, \( p \) has to be a multiple of \( -1 \) and \( \frac{4}{3} \). Because \( -1 \cdot 4 = -4 \) and \( \frac{4}{3} \cdot 3 = 4 \), we know that \( p = -4 \) keeps the proportional relationship.

Hence, \( p \) must be -4 for the system to have no solutions. The constants are non-proportional, \( 0 \), \( \frac{4}{3} \), and \( \frac{3}{2} \), while the coefficients are proportional.

Therefore, the value of \( p \) for which the system has no solution is \( p = -4 \).