For infinitely many solutions, Row3 = m·Row1 + n·Row2 for the coefficients, and the same combination must hold for the constants.

Matching coefficients: for x: \(m+2n=3\); for y: \(m+4n=2\); for z: \(m-n=\lambda\).

Subtracting the x-equation from the y-equation: \((m+4n)-(m+2n)=2-3\Rightarrow 2n=-1\Rightarrow n=-\dfrac12\).

Then \(m+2\left(-\dfrac12\right)=3\Rightarrow m-1=3\Rightarrow m=4\). (Check: \(m+4n=4-2=2\) ✓.)

So \(\lambda=m-n=4-\left(-\dfrac12\right)=\dfrac{9}{2}\).

For the constant term: \(r=2m+6n=2(4)+6\left(-\dfrac12\right)=8-3=5\).

Now check the options with \(\lambda=\dfrac92,\ r=5\): \(2\lambda+r=2\left(\dfrac92\right)+5=9+5=14\), which matches option (B). (The other options do not hold: \(\lambda-2r=4.5-10=-5.5\ne-5\); \(\lambda+2r=4.5+10=14.5\ne14\); \(2\lambda-r=9-5=4\ne5\).)

Therefore the correct relation is \(2\lambda+r=14\).