The number of real values of $$\lambda$$, such that the system of linear equations
$$2x - 3y + 5z = 9$$
$$x + 3y - z = -18$$
$$3x - y + (\lambda^2 - |\lambda|)z = 16$$
has no solutions, is

We need to find the number of real values of $$\lambda$$ for which the system of linear equations has no solution.

The given system can be written as $$ 2x - 3y + 5z = 9 \quad (1) \\ x + 3y - z = -18 \quad (2) \\ 3x - y + (\lambda^2 - |\lambda|)z = 16 \quad (3) $$ and we set $$t = \lambda^2 - |\lambda|$$ to simplify notation.

The determinant of the coefficient matrix is $$ D = \begin{vmatrix} 2 & -3 & 5 \\ 1 & 3 & -1 \\ 3 & -1 & t \end{vmatrix}. $$ Expanding along the first row gives $$ D = 2(3t - 1) - (-3)(t + 3) + 5(-1 - 9) \\ = 2(3t - 1) + 3(t + 3) + 5(-10) \\ = 6t - 2 + 3t + 9 - 50 \\ = 9t - 43. $$

By Cramer's rule, the system is inconsistent precisely when $$D = 0$$ while at least one of $$D_x, D_y, D_z$$ is nonzero. Setting $$D = 0$$ yields $$ 9t - 43 = 0 \quad\Longrightarrow\quad t = \dfrac{43}{9}. $$

Next we replace the first column of the coefficient matrix with the constants to compute $$ D_x = \begin{vmatrix} 9 & -3 & 5 \\ -18 & 3 & -1 \\ 16 & -1 & \dfrac{43}{9} \end{vmatrix}. $$ Expanding this determinant produces $$ D_x = 9\Bigl(\dfrac{43}{3} - 1\Bigr) + 3\Bigl(-18 \cdot \dfrac{43}{9} + 16\Bigr) + 5(18 - 48) \\ = 9 \cdot \dfrac{40}{3} + 3(-86 + 16) + 5(-30) \\ = 120 - 210 - 150 = -240 \neq 0. $$ Since $$D = 0$$ but $$D_x \neq 0$$, the system is indeed inconsistent when $$t = \dfrac{43}{9}$$ and thus has no solution for that value of $$t$$.

We now solve the equation $$ \lambda^2 - |\lambda| = \dfrac{43}{9}. $$ Let $$u = |\lambda|\ge 0$$ so that $$\lambda^2 = u^2$$; the equation becomes $$ u^2 - u = \dfrac{43}{9}, $$ or equivalently $$ 9u^2 - 9u - 43 = 0. $$

By the quadratic formula, $$ u = \dfrac{9 \pm \sqrt{81 + 4 \times 9 \times 43}}{18} = \dfrac{9 \pm \sqrt{1629}}{18}. $$ Since $$\sqrt{1629}\approx 40.36$$, the two values are $$ u_1 = \dfrac{9 + 40.36}{18} \approx 2.74, \quad u_2 = \dfrac{9 - 40.36}{18} \approx -1.74. $$ We discard $$u_2$$ as negative, leaving $$ |\lambda| = u_1 = \dfrac{9 + \sqrt{1629}}{18}. $$

Therefore there are two real values of $$\lambda$$, namely $$ \lambda = +\dfrac{9 + \sqrt{1629}}{18} \quad\text{or}\quad \lambda = -\dfrac{9 + \sqrt{1629}}{18}. $$ Hence, there are 2 real values of $$\lambda$$. The correct answer is Option C: $$2$$.