For a system of 3 linear equations in 3 variables to have infinitely many solutions, the third equation must be a linear combination of the first two, and the system must be consistent.

So let $a(11x+y+\lambda z=-5)+b(2x+3y+5z=3)=(8x-19y-39z=\mu).$

Compare coefficients of $x$ and $y$: $11a+2b=8$ $a+3b=-19$

Solve these:

From $a=-19-3b$ substitute into the first: $11(-19-3b)+2b=8$ $-209-33b+2b=8$ $-31b=217$ $b=-7$

Then $a=-19-3(-7)=2.$

Now compare the coefficient of $z$: $a\lambda+5b=-39$ $2\lambda+5(-7)=-39$ $2\lambda-35=-39$ $2\lambda=-4$ $\lambda=-2.$

Compare the constant terms: $a(-5)+b(3)=\mu$ $2(-5)+(-7)(3)=\mu$ $-10-21=\mu$ $\mu=-31.$

Compute: $\lambda^4-\mu=(-2)^4-(-31)=16+31=47.$

Therefore, the correct option is $\boxed{47}$ i.e. Option C.

Comparison with stored correct answer: Stored correct answer is C, which matches our result.