Let $$\lambda, \mu \in \mathbb{R}$$. If the system of equations
$$3x + 5y + \lambda z = 3$$
$$7x + 11y - 9z = 2$$
$$97x + 155y - 189z = \mu$$
has infinitely many solutions, then $$\mu + 2\lambda$$ is equal to :

For infinitely many solutions, the third equation must be a linear combination of the first two.

Let the third equation = $$\alpha$$(first) + $$\beta$$(second):

$$3\alpha + 7\beta = 97$$ ... (i), $$5\alpha + 11\beta = 155$$ ... (ii), $$\lambda\alpha - 9\beta = -189$$ ... (iii), $$3\alpha + 2\beta = \mu$$ ... (iv)

From (i) and (ii): subtracting gives $$2\beta = 20$$, so $$\beta = 10$$ and $$\alpha = 9$$.

From (iii): $$9\lambda - 90 = -189$$, so $$\lambda = -11$$.

From (iv): $$\mu = 3(9) + 2(10) = 47$$.

$$\mu + 2\lambda = 47 + 2(-11) = 25$$.

The correct answer is Option 2: 25.