`|(0,p-q,p-r),(q-p,0,q-r),(r-p,r-q,0)|=`

Select and write the correct answer from the given alternatives in each of the following: |(0,p-q,q-r),(q-p,0,q-r),(r-p,r-q,0)| is equal to

Choose the correct answer from the bracket.The value of the determinant abs[[0,p-q,p-r],[q-p,0,q-r],[r-p,r-q,0]] is

Let A=|{:(,0,21,r),(,p,q,-r),(,p,-q,r):}|=A A^(T)=I_(3) then |p| is

The lines (p-q)x+(q-r)y+(r-p)=0(q-r)x+(r-p)y+(p-q)=0 (r-p)x+(p-q)y+(q-r)=0

The lines (p-q)x+(q-r)y+(r-p)=0(q-r)x+(r-p)y+(p-q)=0,(x-p)x+(p-q)y+(q-r)=

The lines 1) (p-q)x+(q-r)y+(r-p)=0,2)(q-r)x+(r-p)y+(p-q)=0,3)(r-p)x+(p-q)y+(q-r)=0

Without expanding at any stage, prove that |(p+q, q+r, r+p),(q+r, r+p, p+q),(r+p, p+q, q+r)| = 2|(p,q,r),(q,r,p),(r,p,q)| .