How can I intialise char type array in C

The problem is that you have to declare variables as teh correct type in order to use them:

char arr[3]

Allocates enough space for 3 char items - and in C char is almost certainly ASCII which means it is 8 bits wide, so even on a 64 bit system the arr array occupies only three bytes of memory.
But this:

{"hi","how","are"}

is not three characters, it's three strings, which are much larger.

"hi"   requires 3 bytes: 'h', 'i', and a null terminator.
"how"  requires 4 bytes: 'h', 'o', 'w', and a null terminator.
"are"  requires 4 bytes: 'a', 'r', 'e', and a null terminator.

In addition, you don't store the data in arr itself, arr requires three pointer to char values in order to access the strings, and each pointer needs 8 bytes on a 64 bit system! If it didn't work like that, you couldn't efficiently index through the strings as the "distance" between two strings would not be the same each time.
To use the string initializer, you need to declare arr so that it has enough space to store them, and that means declaring it as an array of pointer to char values:

char* arr[3] = {"hi","how","are"};