For a system of 3 linear equations in 3 variables to have infinitely many solutions, the equations must be dependent.

So the third equation must be a linear combination of the first two, and also the coefficient matrix must be singular.

Write the equations:

Assume the third equation is obtained as $a(2x-y+z)+b(5x+\lambda y+3z)=100x-47y+\mu z$ and similarly on the RHS, $4a+12b=212.$

Compare coefficients of $x$: $2a+5b=100.$

Compare constants: $4a+12b=212.$

Solve for $a,b$.

From $2a+5b=100$ multiply by $2$: $4a+10b=200.$

Subtract from $4a+12b=212$ to get $2b=12 \implies b=6.$

Then $2a+5(6)=100 \implies 2a+30=100 \implies 2a=70 \implies a=35.$

Now compare the coefficient of $y$: $-a+b\lambda=-47.$ Substituting $a=35, b=6$: $-35+6\lambda=-47$ $6\lambda=-12$ $\lambda=-2.$

Compare the coefficient of $z$: $a+3b=\mu.$ So $\mu=35+3(6)=35+18=53.$

Therefore, $\mu-2\lambda=53-2(-2)=53+4=57.$

Hence the correct option is $\boxed{57}$ which is Option C.