`|{:(.^(x)C_(r),,.^(x)C_(r+1),,.^(x)C_(r+2)),(.^(y)C_(r),,.^(y)C_(r+1),,.^(y)C_(r+2)),(.^(z)C_(r),,.^(z)C_(r+1),,.^(z)C_(r+2)):}|` is equal to

Answer

A, B, C, D

Explanation

`Delta =|{:(.^(x)C_(r),,.^(x)C_(r+1),,.^(x)C_(r+2)),(.^(y)C_(r),,.^(y)C_(r+1),,.^(y)C_(r+2)),(.^(z)C_(r),,.^(z)C_(r+1),,.^(z)C_(r+2)):}|`
Applying `C_(2) to C_(2) +C_(1)` we get
`Delta = |{:(.^(x)C_(r),,.^(x+1)C_(r+1),,.^(x)C_(r+2)),(.^(y)C_(r),,.^(y+1)C_(r+1),,.^(y)C_(r+2)),(.^(z)C_(r),,.^(z+1)C_(r+1),,.^(z)C_(r+2)):}|`
so (4) is correct option
In (1) applying `C_(3) to C_(3) +C_(2)` we get
`Delta =|{:(.^(x)C_(r),,.^(x)C_(r+1),,.^(x+1)C_(r+2)),(.^(y)C_(r),,.^(y)C_(r+1),,.^(y+1)C_(r+2)),(.^(z)C_(r),,.^(z)C_(r+1),,.^(z+1)C_(r+2)):}|`
So (3) is correct option
In above applying `C_(2) to C_(2) +C_(1)` we get
`Delta =|{:(.^(x)C_(r),,.^(x+1)C_(r+1),,.^(x+1)C_(r+2)),(.^(y)C_(r),,.^(y+1)C_(r+1),,.^(y+1)C_(r+2)),(.^(z)C_(r),,.^(z+1)C_(r+1),,.^(z+1)C_(r+2)):}|`
So (1) is correct option
In above applying `C_(3) to C_(3)+C_(2)` we get
`Delta = |{:(.^(x)C_(r),,.^(x+1)C_(r+1),,.^(x+2)C_(r+2)),(.^(y)C_(r),,.^(y+1)C_(r+1),,.^(y+2)C_(r+2)),(.^(z)C_(r),,.^(z+1)C_(r+1),,.^(z+2)C_(r+2)):}|`
so (2) is correct option