Prove this equation with the given determinant

Question

Let $$ \begin{vmatrix} a & \sqrt{5} & \sqrt{7} \\ \sqrt{3} & b & \sqrt{7} \\ \sqrt{3} & \sqrt{5} & c \\ \end{vmatrix} =0 $$ with $a\neq\sqrt{3},\, b\neq\sqrt{5},\, c\neq\sqrt{7}.$ Show that $$\frac{a}{a-\sqrt{3}}+\frac{b}{b-\sqrt{5}}+\frac{c}{c-\sqrt{7}}=2$$

Answer 1

I believe that in the proof statement there should be +2 on the RHS. Under that assumption:

$$ \begin{vmatrix} a & \sqrt{5} & \sqrt{7} \\ \sqrt{3} & b & \sqrt{7} \\ \sqrt{3} & \sqrt{5} & c \\ \end{vmatrix} =0 $$ Dividing the first, second and third columns throughout by $\sqrt3, \sqrt5, \sqrt7$ respectively: $$ \begin{vmatrix} a’ & 1 & 1\\ \ 1 & b’ & 1 \\ \ 1 & 1 & c’ \\ \end{vmatrix} =0 $$ [Assuming $a’=\frac{a}{\sqrt3}$ and similarly for others.] As shown here, that implies: $$\frac{1}{1-a’}+ \frac{1}{1-b’}+ \frac{1}{1-c’}=1$$ Substituting the original values leads to the proof of the given statement.