If the system of linear equations
$$2x + y - z = 7$$
$$x - 3y + 2z = 1$$
$$x + 4y + \delta z = k$$, where $$\delta, k \in R$$
has infinitely many solutions, then $$\delta + k$$ is equal to

The system of equations is:

$$2x + y - z = 7 \quad \cdots (1)$$

$$x - 3y + 2z = 1 \quad \cdots (2)$$

$$x + 4y + \delta z = k \quad \cdots (3)$$

For infinitely many solutions, the determinant of the coefficient matrix must be zero, and all sub-determinants formed by replacing columns with the constant column must also be zero.

Step 1: Set $$D = 0$$.

$$D = \begin{vmatrix} 2 & 1 & -1 \\ 1 & -3 & 2 \\ 1 & 4 & \delta \end{vmatrix}$$

$$= 2(-3\delta - 8) - 1(\delta - 2) + (-1)(4 + 3)$$

$$= -6\delta - 16 - \delta + 2 - 7 = -7\delta - 21$$

Setting $$D = 0$$: $$-7\delta - 21 = 0 \implies \delta = -3$$.

Step 2: Find $$k$$ using another determinant condition.

Replace the third column with constants:

$$D_3 = \begin{vmatrix} 2 & 1 & 7 \\ 1 & -3 & 1 \\ 1 & 4 & k \end{vmatrix}$$

$$= 2(-3k - 4) - 1(k - 1) + 7(4 + 3)$$

$$= -6k - 8 - k + 1 + 49 = -7k + 42$$

Setting $$D_3 = 0$$: $$-7k + 42 = 0 \implies k = 6$$.

Step 3: Verify with $$\delta = -3, k = 6$$.

Equation (3) becomes $$x + 4y - 3z = 6$$. Adding equations (1) and (2): $$3x - 2y + z = 8$$. We can verify that equation (3) is a linear combination of (1) and (2): $$(1) - (2)$$ gives $$x + 4y - 3z = 6$$, which matches. $$\checkmark$$

$$\delta + k = -3 + 6 = 3$$

The answer is Option B: 3.