Let $$\theta \in \left(0, \frac{\pi}{2}\right)$$. If the system of linear equations
$$(1 + \cos^2 \theta)x + \sin^2 \theta y + 4\sin 3\theta z = 0$$
$$\cos^2 \theta x + (1 + \sin^2 \theta)y + 4\sin 3\theta z = 0$$
$$\cos^2 \theta x + \sin^2 \theta y + (1 + 4\sin 3\theta)z = 0$$
has a non-trivial solution, then the value of $$\theta$$ is:

We begin by writing the coefficient matrix of the given system

$$ \begin{bmatrix} 1+\cos^2\theta & \sin^2\theta & 4\sin3\theta\\ \cos^2\theta & 1+\sin^2\theta & 4\sin3\theta\\ \cos^2\theta & \sin^2\theta & 1+4\sin3\theta \end{bmatrix}. $$

A homogeneous linear system has a non-trivial solution exactly when the determinant of its coefficient matrix is zero. So we must enforce

$$\det\begin{bmatrix} 1+\cos^2\theta & \sin^2\theta & 4\sin3\theta\\ \cos^2\theta & 1+\sin^2\theta & 4\sin3\theta\\ \cos^2\theta & \sin^2\theta & 1+4\sin3\theta \end{bmatrix}=0.$$

For cleaner algebra let us set

$$a=\cos^2\theta,\qquad b=\sin^2\theta,\qquad c=4\sin3\theta.$$

Because $$\sin^2\theta+\cos^2\theta=1$$ we also have the useful relation $$a+b=1.$$

In these symbols the determinant becomes

$$ \Delta=\det\begin{bmatrix} 1+a & b & c\\ a & 1+b & c\\ a & b & 1+c \end{bmatrix}. $$

To evaluate $$\Delta$$ we perform the row operations $$R_2\to R_2-R_1$$ and $$R_3\to R_3-R_1$$ (these do not change the value of a determinant):

$$ \begin{bmatrix} 1+a & b & c\\ -1 & 1 & 0\\ -1 & 0 & 1 \end{bmatrix}. $$

Now we expand the determinant along the first row. The formula for a $$3\times3$$ determinant is

$$ \det\begin{bmatrix} p & q & r\\ s & t & u\\ v & w & x \end{bmatrix}=p(t x-u w)-q(s x-u v)+r(s w-t v). $$

Applying this with $$(p,q,r)=(1+a,\,b,\,c)$$ we obtain

$$ \Delta=(1+a)\,(1\cdot1-0\cdot0)-b\,((-1)\cdot1-0\cdot(-1))+c\,((-1)\cdot0-1\cdot(-1)). $$

Term by term, we have

$$ 1\cdot1-0\cdot0=1,\qquad (-1)\cdot1-0\cdot(-1)=-1,\qquad (-1)\cdot0-1\cdot(-1)=1. $$

Putting these results back,

$$ \Delta=(1+a)(1)-b(-1)+c(1)=1+a+b+c. $$

Because $$a+b=1$$, substitution gives

$$ \Delta=1+(a+b)+c=1+1+c=2+c. $$

For a non-trivial solution we require $$\Delta=0$$, hence

$$ 2+c=0\quad\Longrightarrow\quad c=-2. $$

Recalling the definition of $$c$$, we find

$$ 4\sin3\theta=-2\quad\Longrightarrow\quad\sin3\theta=-\frac12. $$

Now, $$\theta$$ lies in the interval $$\left(0,\,\dfrac{\pi}{2}\right)$$, so $$3\theta$$ lies in $$\left(0,\,\dfrac{3\pi}{2}\right)$$. In this interval the equation $$\sin\phi=-\dfrac12$$ is satisfied only at

$$ \phi=\frac{7\pi}{6}. $$

Therefore

$$ 3\theta=\frac{7\pi}{6}\quad\Longrightarrow\quad\theta=\frac{7\pi}{18}. $$

Since $$\dfrac{7\pi}{18}$$ indeed lies between $$0$$ and $$\dfrac{\pi}{2}$$, this is the unique admissible value of $$\theta$$.

Hence, the correct answer is Option C.