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

Step 1: Find the value of k from the first system of equations

The first system of linear equations is given by:

$$x + y + kz = 2$$

$$2x + 3y - z = 1$$

$$3x + 4y + 2z = k$$

For a system of linear equations to have infinitely many solutions, the determinant of the coefficient matrix, $$\Delta$$, must be equal to zero.

$$\Delta = \begin{vmatrix} 1 & 1 & k \\ 2 & 3 & -1 \\ 3 & 4 & 2 \end{vmatrix} = 0$$

Expanding the determinant along the first row:

$$\Delta = 1(3(2) - (-1)(4)) - 1(2(2) - (-1)(3)) + k(2(4) - 3(3)) = 0$$

$$\Delta = 1(6 + 4) - 1(4 + 3) + k(8 - 9) = 0$$

$$\Delta = 10 - 7 - k = 0$$

$$3 - k = 0$$

$$k = 3$$

We can also verify this by adding the first two equations:

$$(x + y + kz) + (2x + 3y - z) = 2 + 1$$

$$3x + 4y + (k-1)z = 3$$

Comparing this with the third equation $$3x + 4y + 2z = k$$, we see that for infinitely many solutions, the coefficients of $$z$$ and the constant terms must match:

$$k - 1 = 2 \implies k = 3$$

$$k = 3$$

Step 2: Substitute k = 3 into the second system of equations

The second system of equations is given by:

$$(k+1)x + (2k-1)y = 7$$

$$(2k+1)x + (k+5)y = 10$$

Substituting $$k = 3$$ into these equations:

$$L_1 : (3+1)x + (2(3)-1)y = 7 \implies 4x + 5y = 7$$

$$L_2 : (2(3)+1)x + (3+5)y = 10 \implies 7x + 8y = 10$$

Step 3: Calculate the point of intersection

To find the unique point of intersection, we can solve the system of linear equations:

Equation 1: $$4x + 5y = 7$$

Equation 2: $$7x + 8y = 10$$

Multiplying Equation 1 by 7 and Equation 2 by 4 to eliminate $$x$$:

$$28x + 35y = 49$$

$$28x + 32y = 40$$

Subtracting the second equation from the first:

$$(28x - 28x) + (35y - 32y) = 49 - 40$$

$$3y = 9$$

$$y = 3$$

Substituting $$y = 3$$ back into Equation 1:

$$4x + 5(3) = 7$$

$$4x + 15 = 7$$

$$4x = 7 - 15$$

$$4x = -8$$

$$x = -2$$

The two lines intersect at a single unique point, which is $$(-2, 3)$$.

Conclusion:

The system has a unique solution at the point of intersection $$(-2, 3)$$ which satisfy the condition $$x + y = 1$$.