Three points $X, Y,Z$ are on a striaght line such that $XY = 10$ and $XZ = 3$. What is the product of all possible values of $YZ$?
To solve this problem, we need to consider the relative positions of the points \(X\), \(Y\), and \(Z\) on a straight line. We are given that \(XY = 10\) and \(XZ = 3\). We need to find the product of all possible values of \(YZ\).

1. **Case 1: \(X\) lies between \(Y\) and \(Z\)**

   In this case, the distance \(YZ\) is the sum of \(XY\) and \(XZ\):
   \[
   YZ = XY + XZ = 10 + 3 = 13
   \]

2. **Case 2: \(Y\) lies between \(X\) and \(Z\)**

   In this case, the distance \(YZ\) is the difference between \(XY\) and \(XZ\):
   \[
   YZ = XY - XZ = 10 - 3 = 7
   \]

3. **Case 3: \(Z\) lies between \(X\) and \(Y\)**

   This case is essentially the same as Case 2, where \(YZ = 7\). Therefore, we do not need to consider it separately.

Now, we have two possible values for \(YZ\): 13 and 7. The product of these values is:
\[
13 \times 7 = 91
\]

The final answer is \(\boxed{91}\)