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

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

Write the equations:

Assume the third equation is obtained as a linear combination of the first two: $a(x+2y-3z=2)+b(2x+\lambda y+5z=5)=14x+3y+\mu z=33$

Comparing coefficients, we get: $a+2b=14 \quad ...(1)$ $2a+\lambda b=3 \quad ...(2)$ $-3a+5b=\mu \quad ...(3)$ $2a+5b=33 \quad ...(4)$

Solve for $a$ and $b$ using (1) and (4):

From (1): $a=14-2b$

Substitute into (4): $2(14-2b)+5b=33$ $28-4b+5b=33$ $b=5$

Then $a=14-2(5)=4$

Now use (2) to find $\lambda$: $2a+\lambda b=3$ $2(4)+5\lambda=3$ $8+5\lambda=3$ $5\lambda=-5$ $\lambda=-1$

Use (3) to find $\mu$: $\mu=-3a+5b$ $\mu=-3(4)+5(5)$ $\mu=-12+25=13$

Therefore, $\lambda+\mu=-1+13=12$

Hence the correct option is: $\boxed{12}$ which is Option C.