If the system of equations
$$2x + \lambda y + 3z = 5$$
$$3x + 2y - z = 7$$
$$4x + 5y + \mu z = 9$$
has infinitely many solutions, then $$(\lambda^2 + \mu^2)$$ is equal to :

For a system of three linear equations to possess infinitely many solutions,
the coefficient matrix $$A$$ must be singular (determinant $$0$$) and the augmented matrix must have the same rank as $$A$$ (that rank will then be $$2$$).

Write the coefficient matrix and its determinant:

$$A=\begin{vmatrix}2 & \lambda & 3\\ 3 & 2 & -1\\ 4 & 5 & \mu\end{vmatrix}$$

Expanding along the first row,

$$\begin{aligned} \det A &= 2\begin{vmatrix}2 & -1\\ 5 & \mu\end{vmatrix} -\lambda\begin{vmatrix}3 & -1\\ 4 & \mu\end{vmatrix} +3\begin{vmatrix}3 & 2\\ 4 & 5\end{vmatrix}\\[4pt] &= 2(2\mu+5)-\lambda(3\mu+4)+3(15-8)\\[4pt] &= 4\mu+10-\lambda(3\mu+4)+21\\[4pt] &= 4\mu+31-\lambda(3\mu+4). \end{aligned}$$

Setting $$\det A = 0$$ gives

$$\lambda(3\mu+4)=4\mu+31 \quad -(1)$$

Because $$\det A = 0$$, the three rows are linearly dependent.
Assume the third row is a linear combination of the first two:

$$R_3 = pR_1 + qR_2$$

This yields four scalar equations (for the coefficients of $$x,\,y,\,z$$ and the constants):

$$\begin{aligned} 2p+3q &= 4 \quad -(2)\\ \lambda p + 2q &= 5 \quad -(3)\\ 3p - q &= \mu \quad -(4)\\ 5p + 7q &= 9 \quad -(5) \end{aligned}$$

Solve equations $$(2)$$ and $$(3)$$ for $$p,\,q$$.
From $$(2):$$ $$p=\dfrac{4-3q}{2}.$$
Insert this into $$(3):$$ $$5 = \lambda\left(\dfrac{4-3q}{2}\right)+2q \;\;\Longrightarrow\;\; (-3\lambda+4)q = 10-4\lambda.$$

Set $$D = 4-3\lambda$$ (note $$D\neq0$$ for rank $$2$$). Then

$$q=\dfrac{10-4\lambda}{D},\qquad p=\dfrac{-7}{D}.$$

Check the constant equation $$(5)$$ for consistency:

$$5p+7q \;=\; 5\!\left(\dfrac{-7}{D}\right)+7\!\left(\dfrac{10-4\lambda}{D}\right) = \dfrac{35-28\lambda}{D}.$$

For infinite solutions this must equal $$9$$:

$$\dfrac{35-28\lambda}{D}=9 \;\;\Longrightarrow\;\; 35-28\lambda = 9(4-3\lambda) = 36-27\lambda.$$

Simplifying gives $$\lambda = -1.$$

Now substitute $$\lambda=-1$$ into $$(1)$$ to find $$\mu$$:

$$(-1)(3\mu+4)=4\mu+31 \;\;\Longrightarrow\;\; -3\mu-4 = 4\mu+31 \;\;\Longrightarrow\;\; 7\mu = -35 \;\;\Longrightarrow\;\; \mu = -5.$$

Finally,

$$(\lambda^2+\mu^2)=(-1)^2+(-5)^2 = 1+25 = 26.$$

Hence the required value is $$26$$, corresponding to Option C.