Determinant in symmetrical form

$\begin{aligned}\left | \begin{array}{ccc} ap & a & p \\ bq & b & q \\ cr & c & r \\ \end{array} \right |&=abcpqr\left | \begin{array}{ccc} 1 &\frac{1}{p} &\frac{1}{a} \\1 &\frac{1}{q} &\frac{1}{b} \\1 &\frac{1}{r} &\frac{1}{c} \\ \end{array} \right |\\&=\left | \begin{array}{ccc} 1 &qr & bc\\1 &pr &ca \\1 &pq & ab \\ \end{array} \right |\\&=\left | \begin{array}{ccc} 1 &qr & bc+qr+1\\1 &pr &ca+pr+1 \\1 &pq & ab+pq+1 \\ \end{array} \right |\\&=\left | \begin{array}{ccc} 1 &qr & 0\\1 &pr &0 \\1 &pq & 0\\ \end{array} \right |\\&\huge\color{#3D99F6}{=\boxed{0}}\end{aligned}$

$\left| \begin{array}{ccc} ap & a & p \\ bq & b & q \\ cr & c & r \end{array} \right|$ $= apbr+aqcr+pbqc-pbcr-apcq-abqr$ $= ab(pr - qr) + ac(qr - pq) + bc(pq - pr)$ $= ab(bc-ca) + ac(ab-bc) + bc(ac-ab)$ $= abc\left((b-a)+(a-c)+(c-b)\right) = \boxed{0}$

P.S. Notice that the $-1$ is not necessary. It can be replaced with any number.

Best Solution :

let b=c=0

then you have from relations

p=q=r=i= root(-1)

(nothing has been told about them being real or not any way)

now put in determinant to get

|ai a i|
|bi b i |

|ci c i|

which is obviously 0 (as second column * i = 1st column)

Did same as Ariel Gershon