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Let the system of linear equations
$x + y + kz = 2$ ,
$2 x + 3 y - z = 1$ ,
$3 x + 4 y + 2 z = k$ .
Have infinitely many solutions. Then the system
$(k + 1) x + (2 k - 1) y = 7,$
$(2 k + 1) x + (k + 5) y = 10,$ has:

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infinitely many solutions

unique solutions satisfying $x - y = 1$

unique solutions satisfying $x + y = 1$

No solutions

answer is B.

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Detailed Solution

$Δ = |\begin{matrix} 1 & 1 & k \\ 2 & 3 & - 1 \\ 3 & 4 & 2 \end{matrix}| = 0 \Rightarrow 10 - 7 - k = 0 \Rightarrow k = 3$
For k=3, clearly addition of first 2 equations given the third one

$\begin{array}{l} 4 x + 5 y = 7 \\ 7 x + 8 y = 10 \\ x = - 2, y = 3 \end{array}$

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