The value of $$k \in R$$, for which the following system of linear equations
$$3x - y + 4z = 3$$
$$x + 2y - 3z = -2$$
$$6x + 5y + kz = -3$$
has infinitely many solutions, is:

For infinitely many solutions,

$$D=0$$ where

$$D$$ is the determinant of the coefficient matrix.

Thus,

$$D=\begin{vmatrix} 3&-1&4\\ 1&2&-3\\ 6&5&k \end{vmatrix}$$

Expanding along the first row,

$$D= 3\begin{vmatrix} 2&-3\\ 5&k \end{vmatrix} -(-1)\begin{vmatrix} 1&-3\\ 6&k \end{vmatrix} +4\begin{vmatrix} 1&2\\ 6&5 \end{vmatrix}$$

$$=3(2k+15)+(k+18)+4(5-12)$$

$$=6k+45+k+18-28$$

$$=7k+35$$

For infinitely many solutions,

$$7k+35=0$$

$$7k=-35$$

$$k=-5$$

Now verify consistency.

Let

$$R_3=\alpha R_1+\beta R_2$$

Comparing coefficients,

$$3\alpha+\beta=6$$

$$-\alpha+2\beta=5$$

Solving,

$$\beta=3,\qquad \alpha=1$$

Now check constants:

$$1(3)+3(-2)=-3$$ and $$1(4)+3(-3)=-5$$

Hence third equation is exactly $$R_1+3R_2$$

Therefore system is dependent and consistent, so it has infinitely many solutions.

Hence, the required value is $$\boxed{-5}$$