For a system of 3 linear equations in 3 variables to have infinitely many solutions, we need:

$\operatorname{rank}(A)=\operatorname{rank}([A|B])<3$

where $A=\begin{pmatrix}2&\lambda&3\\3&2&-1\\4&5&\mu\end{pmatrix},\qquad B=\begin{pmatrix}5\\7\\9\end{pmatrix}.$

First, for rank to be less than 3, we must have $\det(A)=0.$

Compute: $\det(A)=\begin{vmatrix}2&\lambda&3\\3&2&-1\\4&5&\mu\end{vmatrix}$

Expanding along the first row, $\det(A)=2\begin{vmatrix}2&-1\\5&\mu\end{vmatrix}-\lambda\begin{vmatrix}3&-1\\4&\mu\end{vmatrix}+3\begin{vmatrix}3&2\\4&5\end{vmatrix}.$

Now, $\begin{vmatrix}2&-1\\5&\mu\end{vmatrix}=2\mu+5,$ $\begin{vmatrix}3&-1\\4&\mu\end{vmatrix}=3\mu+4,$ $\begin{vmatrix}3&2\\4&5\end{vmatrix}=15-8=7.$

Therefore, $\det(A)=2(2\mu+5)-\lambda(3\mu+4)+3(7)$ $=4\mu+10-\lambda(3\mu+4)+21$ $=4\mu+31-\lambda(3\mu+4).$

So, $4\mu+31-\lambda(3\mu+4)=0. \qquad (1)$

For infinitely many solutions, the third equation must be a linear combination of the first two, and the constants must satisfy the same relation.

Let $a(2x+\lambda y+3z=5)+b(3x+2y-z=7)=(4x+5y+\mu z=9).$

Comparing coefficients: $2a+3b=4 \qquad (2)$ $\lambda a+2b=5 \qquad (3)$ $3a-b=\mu \qquad (4)$ $5a+7b=9 \qquad (5)$

Solve for $a,b$ using (2) and (5):

From (2): $2a+3b=4$ From (5): $5a+7b=9$

Multiply first by 5: $10a+15b=20$ Multiply second by 2: $10a+14b=18$

Subtract: $b=2.$

Then from (2): $2a+3(2)=4$ $2a+6=4$ $2a=-2$ $a=-1.$

Now use (3) and (4):

From (3): $\lambda(-1)+2(2)=5$ $-\lambda+4=5$ $\lambda=-1.$

From (4): $\mu=3(-1)-2=-3-2=-5.$

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

Therefore the correct option is $\boxed{\text{B: }26}.$