For the system to have infinitely many solutions, the three equations must be dependent and consistent.

So, the third equation must be a linear combination of the first two.

Let us check whether $a(3x+5y+\lambda z=3)+b(7x+11y-9z=2)=(97x+155y-189z=\mu).$

Comparing coefficients of $x$ and $y$: $3a+7b=97$ $5a+11b=155$

Solve these two equations.

From $3a+7b=97 \quad ...(1)$ $5a+11b=155 \quad ...(2)$

Multiply (1) by $5$ and (2) by $3$: $15a+35b=485$ $15a+33b=465$

Subtract: $2b=20 \Rightarrow b=10$

Put in (1): $3a+70=97 \Rightarrow 3a=27 \Rightarrow a=9$

Now compare the coefficient of $z$.

Since the third equation must be $9\times$(first) $+10\times$(second), $9\lambda +10(-9)=-189$ $9\lambda-90=-189$ $9\lambda=-99$ $\lambda=-11$

Compare constants to get $\mu$: $\mu=9(3)+10(2)=27+20=47$

Now compute: $\mu+2\lambda=47+2(-11)=47-22=25$

Therefore, the correct option is $\boxed{25}$ i.e. Option B.