The values of $$m, n$$, for which the system of equations $$x + y + z = 4$$, $$2x + 5y + 5z = 17$$, $$x + 2y + mz = n$$ has infinitely many solutions, satisfy the equation:

The system of equations is:

$$x + y + z = 4$$ ... (1)

$$2x + 5y + 5z = 17$$ ... (2)

$$x + 2y + mz = n$$ ... (3)

For infinitely many solutions, the system must be consistent and the determinant of the coefficient matrix must be zero.

Find the determinant.

$$ D = \begin{vmatrix} 1 & 1 & 1 \\ 2 & 5 & 5 \\ 1 & 2 & m \end{vmatrix} $$

$$ D = 1(5m - 10) - 1(2m - 5) + 1(4 - 5) $$

$$ D = 5m - 10 - 2m + 5 - 1 = 3m - 6 $$

Setting $$D = 0$$: $$3m - 6 = 0 \Rightarrow m = 2$$.

For infinitely many solutions, we also need consistency.

With $$m = 2$$, equation (3) becomes: $$x + 2y + 2z = n$$.

From equations (1) and (2): Subtract 2×(1) from (2): $$3y + 3z = 9 \Rightarrow y + z = 3$$.

From (1): $$x = 4 - (y + z) = 4 - 3 = 1$$. So $$x = 1$$.

Substituting into (3): $$1 + 2y + 2z = n \Rightarrow 1 + 2(y + z) = n \Rightarrow 1 + 6 = n \Rightarrow n = 7$$.

Check which equation is satisfied.

$$m = 2, n = 7$$.

Option (1): $$m^2 + n^2 - mn = 4 + 49 - 14 = 39$$. ✓

The correct answer is Option (1): $$m^2 + n^2 - mn = 39$$.