Distance from point to plane

1. Direction Vector of $$L_1$$
Line $$L_1$$ is parallel to the vector $$\vec{v}_1 = 2\hat{i} - \hat{j} + 3\hat{k}$$. It passes through point $$P(1, 2, 1)$$.

2. Direction Vector of $$L_2$$
Line $$L_2$$ is the intersection of planes $$x + 2y - z = 3$$ and $$2x - y + 3z = 1$$. Its direction vector $$\vec{v}_2$$ is perpendicular to both normal vectors $$\vec{n}_1 = (1, 2, -1)$$ and $$\vec{n}_2 = (2, -1, 3)$$:
$$\vec{v}_2 = \vec{n}_1 \times \vec{n}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & -1 \\ 2 & -1 & 3 \end{vmatrix} = 5\hat{i} - 5\hat{j} - 5\hat{k}$$
We can simplify this parallel vector to $$(1, -1, -1)$$.

3. Equation of Plane $$\pi$$
Plane $$\pi$$ contains $$L_1$$ and is parallel to $$L_2$$. Its normal vector $$\vec{n}_{\pi}$$ is:
$$\vec{n}_{\pi} = \vec{v}_1 \times \vec{v}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -1 & 3 \\ 1 & -1 & -1 \end{vmatrix} = 4\hat{i} + 5\hat{j} - \hat{k}$$
Since the plane contains point $$P(1, 2, 1)$$, its equation is:
$$4(x - 1) + 5(y - 2) - 1(z - 1) = 0 \implies 4x + 5y - z - 13 = 0$$

4. Distance from $$Q(2, 3, 4)$$ to Plane $$\pi$$
Using the point-to-plane distance formula:
$$\text{Distance} = \frac{|4(2) + 5(3) - (4) - 13|}{\sqrt{4^2 + 5^2 + (-1)^2}} = \frac{|8 + 15 - 4 - 13|}{\sqrt{16 + 25 + 1}} = \frac{6}{\sqrt{42}}$$
Note: $$\frac{6}{\sqrt{42}} = \frac{\sqrt{36}}{\sqrt{42}} = \sqrt{\frac{6}{7}}$$.