Solve if you can

We have $a+b+c=210$

From rest equations we find (by adding the rest)

$2(a+b+c)+x(a+b+c)=690$ $x=\frac{9}{7}$

We can find from first equation

$a+b=210-c$ $b+c=210-a$ $c+a=210-b$

Now putting the above given value in the parameter $x$ equations

$a(x-1)=10$ $b(x-1)=20$ $c(x-1)=30$

From these

$a:b:c=1:2:3$

From first $a=35, b=70, c=105$

Now we have to find

$490x(a^{-1}+b^{-1}+c^{-1})$ $490\times\frac{9}{7}(\frac{1}{35}+\frac{1}{70}+\frac{1}{105})$ $=33$

$ax+b+c=220=210+10=a+b+c+10 \implies a(x-1)=10$ by the same approach using the other 2 equations you get $b(x-1)=20$ and $c(x-1)=30$ so $b(x-1)=2a(x-1) \implies b=2a$ and $c(x-1)=3a(x-1)\implies c=3a$ so $6a=210 \implies a=35\implies b=70$ and $c=105$ . Now that we have the values of $a$ , $b$ , and $c$ , we can now plug them in any equation with $x$ in it: $35x+70+105=220\implies 35x=45\implies x=\frac{9}{7}$ then $490\cdot\frac{9}{7}(\frac{1}{35}+\frac{1}{70}+\frac{1}{105})$ and solve it which gives 33. This isn't really a number theory question, just simple algebra(maybe not "that" simple) and it certainly isn't level 3.