Fa,b,c" are in "AP" find "|[x+1,x+2,x+a],[x+2,x+3,x+b],[x+3,x+4,x+c]|

If a,b,c are in AP show that |[x+1,x+2,x+a],[x+2,x+3,x+b],[x+3,x+4,x+c]|=0

If a,b,c are in AP show that |[x+1,x+2,x+a],[x+2,x+3,x+b],[x+3,x+4,x+c]|=0

If a,b,c, are in A.P. then |(x+1,x+2,x+a),(x+2,x+3,x+b),(x+3,x+4,x+c)|=

if a,b,c are in A.P. show that : | [ x+1, x+2,x+a],[x+2,x+3,x+b],[x+3,x+4,x+c ] |=0

If a, b , c are in A.P show that |[x+1, x+2, x+a],[ x+2, x+3, x+b],[ x+3, x+4, x+c]|=0

If a,b,c are A.P then |{:(x+1,x+2,x+a),(x+2,x+3,x+b),(x+3,x+4,x+c):}|=0

if a, b,c are in A.P. then what is the value of: |[x+1,x+2,x+a],[x+2,x+3,x+b],[x+3,x+4,x+c]| ?

If a,b,c are in A.P., show that : |(x+1,x+2,x+a),(x+2,x+3,x+b),(x+3,x+4,x+c)|=0

If a,b,c are in A.P., then the determinant |[x+2, x+3, x+2a],[x+3,x+4,x+2b],[x+4,x+5,x+2c]| is

Given a,b,c are in A.P. Then determinant : |(x+1,x+2,x+a),(x+2,x+3,x+b),(x+3,x+4,x+c)| in its simplified form is :