Ratio!

(This solution is taught by @Takeda Shigenori)

From the question, we can get $\begin{cases} -ka+ \sqrt{15}b+ \sqrt{35}c=0 \\ \sqrt{15}a-kb+ \sqrt{21}c=0 \\ \sqrt{35}a+\sqrt{21}b-kc=0 \end{cases}$ Which means that $\begin{pmatrix} -k&\sqrt{15}&\sqrt{35} \\ \sqrt{15}&-k&\sqrt{21} \\ \sqrt{35}& \sqrt{21}&-k \end{pmatrix} \begin{pmatrix} a\\b\\c \end{pmatrix} =\begin{pmatrix} 0\\0\\0 \end{pmatrix}$ By the matrix equation above, we can obviously get that $a=b=c=0$ is one of the solution. But the question said that $a,b,c>0$ . Hence, there exist more than $\large1$ solution for the matrix equation above.

This imply that the matrix $\begin{pmatrix} -k&\sqrt{15}&\sqrt{35} \\ \sqrt{15}&-k&\sqrt{21} \\ \sqrt{35}& \sqrt{21}&-k \end{pmatrix}$ does not have inverse matrix. This is because if the matrix above have inverse matrix, then the matrix equation above have only one solution, which has been proved that is impossible. As the matrix above don't have inverse matrix, its determinant is $0$ .

By the formula of finding determinant of $3\times 3$ matrix, we can get $-k^3+\sqrt {15} \sqrt{21} \sqrt{35}+\sqrt {15} \sqrt{21} \sqrt{35}-(35k+21k+15k) = 0$ $k^3-71k-210=0$ $\large k^3-71k+1=211$