1. **State the problem:** Find the equation of the plane containing the lines $L_1$ and $L_2$.\n\n2. **Write parametric equations of the lines:**\n$L_1:\ x = t,\ y = 2 - t,\ z = 3$\n$L_2:\ x = -1 + 2t,\ y = 3 + 2t,\ z = 3t$\n\n3. **Find direction vectors of the lines:**\nFor $L_1$, direction vector $\vec{d_1} = (1, -1, 0)$ since $x$ increases by 1, $y$ decreases by 1, $z$ constant.\nFor $L_2$, direction vector $\vec{d_2} = (2, 2, 3)$.\n\n4. **Find a point on each line:**\nFor $L_1$, at $t=0$, point $P_1 = (0, 2, 3)$.\nFor $L_2$, at $t=0$, point $P_2 = (-1, 3, 0)$.\n\n5. **Find vector between points on the two lines:**\n$\vec{P_1P_2} = P_2 - P_1 = (-1 - 0, 3 - 2, 0 - 3) = (-1, 1, -3)$.\n\n6. **Find normal vector to the plane:**\nThe plane contains both direction vectors $\vec{d_1}$ and $\vec{d_2}$, so normal vector $\vec{n} = \vec{d_1} \times \vec{d_2}$.\nCalculate cross product:\n$$\vec{n} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & -1 & 0 \\ 2 & 2 & 3 \end{vmatrix} = \mathbf{i}((-1)(3) - 0(2)) - \mathbf{j}(1(3) - 0(2)) + \mathbf{k}(1(2) - (-1)(2))$$\n$$= \mathbf{i}(-3) - \mathbf{j}(3) + \mathbf{k}(2 + 2) = (-3, -3, 4)$$\n\n7. **Write equation of the plane:**\nUsing point $P_1 = (0, 2, 3)$ and normal vector $\vec{n} = (-3, -3, 4)$, plane equation is:\n$$-3(x - 0) - 3(y - 2) + 4(z - 3) = 0$$\nSimplify:\n$$-3x - 3y + 6 + 4z - 12 = 0$$\n$$-3x - 3y + 4z - 6 = 0$$\nDivide entire equation by $-3$ to simplify:\n$$\cancel{-3}x + \cancel{-3}y - \frac{4}{3}z + 2 = 0$$\nCorrecting division carefully:\nDividing by $-3$:\n$$x + y - \frac{4}{3}z + 2 = 0$$\nOr multiply both sides by 3 to avoid fractions:\n$$3x + 3y - 4z - 6 = 0$$\n\n**Final answer:** The equation of the plane is $3x + 3y - 4z - 6 = 0$.