Find the values maked by letters in Equal matrix (3,4,1,*,y,z,19,28,5,2,49,7) equals (p,4,1,6,12,24,19,28,q,z,7) find the value

Let the first matrix be $A$ and the second matrix be $B$.

Matrix $A$ has 12 elements.
Matrix $B$ has 11 elements.
For two matrices to be equal, they must have the same dimensions and thus the same number of elements. Since the number of elements is different, the given matrices cannot be equal as stated.

Assuming the problem intends for element-wise comparison up to the length of the shorter matrix (11 elements), and then addressing the remaining element:

• Comparing the first elements: $3 = p \implies p = 3$
• Comparing the fourth elements: $* = 6$
• Comparing the fifth elements: $y = 12$
• Comparing the sixth elements: $z = 24$
• Comparing the ninth elements: $5 = q \implies q = 5$
• Comparing the tenth elements: $2 = z$
• Comparing the eleventh elements: $49 = 7$ (This is a direct numerical contradiction, as $49 \neq 7$)

From the comparison of the sixth elements, we found $z = 24$.
From the comparison of the tenth elements, we found $z = 2$.
A single variable $z$ cannot simultaneously hold two different values (24 and 2). This is a contradiction.
Additionally, the eleventh elements 49 and 7 are not equal, which further confirms the matrices cannot be equal.

Conclusion: Due to the different number of elements and the direct contradictions for the value of $z$ and the eleventh elements, the given matrices cannot be equal as stated. Therefore, a consistent set of values for all letters cannot be found.

However, if we consider the values that are consistently defined before the contradictions arise, we can state:

• The value marked by $*$ is $\boxed{ = 6}$.
• The value of $y$ is $\boxed{y = 12}$.
• The value of $p$ is $\boxed{p = 3}$.
• The value of $q$ is $\boxed{q = 5}$. The value of $z$ cannot be uniquely determined due to the contradiction ($z=24$ and $z=2$).