If the line of intersection of the planes $$ax + by = 3$$ and $$ax + by + cz = 0$$, $$a > 0$$ makes an angle $$30^\circ$$ with the plane $$y - z + 2 = 0$$, then the direction cosines of the line are

We need to find the direction cosines of the line of intersection of the planes $$ax + by = 3$$ and $$ax + by + cz = 0$$, given that this line makes an angle of $$30°$$ with the plane $$y - z + 2 = 0$$.

Plane 1: $$ax + by + 0z = 3$$ has normal $$\vec{n_1} = (a, b, 0)$$.

Plane 2: $$ax + by + cz = 0$$ has normal $$\vec{n_2} = (a, b, c)$$.

The direction of the line of intersection is $$\vec{d} = \vec{n_1} \times \vec{n_2}$$:

$$\vec{d} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ a & b & 0 \\ a & b & c \end{vmatrix} = \vec{i}(bc - 0) - \vec{j}(ac - 0) + \vec{k}(ab - ab) = (bc, -ac, 0)$$ $$\vec{d} = c(b, -a, 0)$$

Since $$c \neq 0$$ (otherwise the planes would be parallel or identical), the direction is proportional to $$(b, -a, 0)$$.

The plane $$y - z + 2 = 0$$ has normal $$\vec{n_3} = (0, 1, -1)$$.

The angle between a line and a plane is the complement of the angle between the line and the normal. If the line makes angle $$30°$$ with the plane:

$$\sin 30° = \frac{|\vec{d} \cdot \vec{n_3}|}{|\vec{d}||\vec{n_3}|}$$ $$\sin 30° = \frac{|0 \cdot b + (-a)(1) + 0 \cdot (-1)|}{\sqrt{a^2 + b^2} \cdot \sqrt{2}} = \frac{|a|}{\sqrt{a^2 + b^2} \cdot \sqrt{2}}$$ $$\frac{1}{2} = \frac{a}{\sqrt{a^2 + b^2} \cdot \sqrt{2}}$$

(Since $$a > 0$$, $$|a| = a$$.)

$$\sqrt{a^2 + b^2} \cdot \sqrt{2} = 2a$$ $$2(a^2 + b^2) = 4a^2$$ $$2b^2 = 2a^2$$ $$b^2 = a^2 \implies b = \pm a$$

The direction is $$(b, -a, 0)$$. With $$b = a$$: direction is $$(a, -a, 0)$$, giving direction cosines $$\left(\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}, 0\right)$$.

With $$b = -a$$: direction is $$(-a, -a, 0)$$, giving direction cosines $$\left(-\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}, 0\right)$$.

Comparing with the options, $$\left(\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}, 0\right)$$ matches Option B.

The correct answer is Option B: $$\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}, 0$$.