Need help in understanding the solution to the C exercise involving pointers - c

Question

I need help in understanding how we got the values in the table below for Loc3 and Loc4.
When I was making a table on my own I arrived to totally different entries for those columns.
Thank you!
int x = 42; /* x is at address 100 */
int y = 13; /* y is at address 104 */
int *p; /* p is at address 108 */
int **p2; /* p2 is at address 112 */
/* Location 1 */
p = &y;
p2 = &p;
/* Location 2 */
*p2 = &x;
**p2 = 11;
/* Location 3 */
*p = 12;
/* Location 4 */

Answer 1

For instance, x at loc3 becomes 11 because you set **p2 to 11, which modifies the value at that memory location. (double star is a pointer to a pointer). Ampersand gets the address.
To elaborate:
*p2 = &x;
**p2 = 11;
In Loc2 you set p2 = &p, which means p2 is now pointing to the address of p, which is 108.
But now in Loc3, you set what p2 is pointing to to the address of x. In other words, since p2 was pointing to the address of p, now you're saying that p should instead point to the address of x (which is also why p becomes 100).
Then **p2 modifies the value at that address of x to be 11 (through p), hence loc3's x value becomes 11.

Answer 2

Location 1:
int x = 42;
int y = 13;
int *p;
int **p2;
p or p2 don't point anywhere.
Location 2:
p = &y;
p points to y.
p2 = &p;
p2 points to p.
No changes to x or y.
Location 3:
*p2 = &x;
Since p2 points to p, dereferencing p2 and assigning a value to it changes p to point to x. Same as doing p = &x.
**p2 = 11;
Dereference once to get to p, dereference again to get to x, and assign 11 to it. Same as doing: x = 11 or *p == 11.
No change to y or p2.
Location 4:
*p = 12;
Dereference p to get to x and assing 12 to it. No change to y or p2 or p.

Answer 3

When you start out, p and p2 are uninitialized and contain indeterminate values, hence the ?? in both entries under Loc 1.
p = &y;
assigns the location of y (104) to p.
p2 = &p;
assigns the location of p (108) to p2. Note that the type of the expression &p is char **, which matches the type of the variable p2. So after these two statements, all of the following are true:
p2 == &p == 108 // all expressions have type char **
*p2 == p == &y == 104 // all expressions have type char *
**p2 == *p == y == 13 // all expressions have type char
x == 42
Next we execute
*p2 = &x;
From above we see that *p2 is equivalent to p, so this statement assigns the address of x (100) to p, so now we have
p2 == &p == 108
*p2 == p == &x == 100
**p2 == *p == x == 42
y == 13
Next we execute
**p2 = 11;
**p2 is equivalent to *p, which is equivalent to x, so we wind up assigning the value 11 to x:
p2 == &p == 108
*p2 == p == &x == 100
**p2 == *p == x == 11
y == 13
Finally we have
*p = 12;
*p is equivalent to x, so we're assigning the value 12 to x, leaving us with:
p2 == &p == 108
*p2 == p == &x == 100
**p2 == *p == x == 12
y == 13