The teacher hasn't uploaded a solution for this question yet.
Let the general points on the two lines be $P(2\lambda+1, 3\lambda+2, 4\lambda+3)$ and $Q(5\mu+4, 2\mu+1, \mu)$. For intersection, $P=Q$: $$2\lambda+1 = 5\mu+4 \implies 2\lambda - 5\mu = 3$$ $$3\lambda+2 = 2\mu+1 \implies 3\lambda - 2\mu = -1$$ Solving these, multiply first by 3 and second by 2: $6\lambda - 15\mu = 9$ and $6\lambda - 4\mu = -2$. Subtracting gives $-11\mu = 11 \implies \mu = -1$. Substituting $\mu = -1$ into $3\lambda - 2(-1) = -1 \implies 3\lambda = -3 \implies \lambda = -1$. The point is $(-1, -1, -1)$.
The line passes through $\vec{a} = -\hat{i} - \hat{j} - \hat{k}$ and is parallel to $\vec{b} = 3\hat{i} + 2\hat{j} - 8\hat{k}$. The vector equation is $\vec{r} = \vec{a} + \lambda\vec{b}$: $$\vec{r} = (-\hat{i} - \hat{j} - \hat{k}) + \lambda(3\hat{i} + 2\hat{j} - 8\hat{k})$$
Using the point $(x_1, y_1, z_1) = (-1, -1, -1)$ and direction ratios $(a, b, c) = (3, 2, -8)$, the equation is: $$\frac{x+1}{3} = \frac{y+1}{2} = \frac{z+1}{-8}$$
Final Answer: Vector: \vec{r} = (-\hat{i} - \hat{j} - \hat{k}) + \lambda(3\hat{i} + 2\hat{j} - 8\hat{k}); Cartesian: \frac{x+1}{3} = \frac{y+1}{2} = \frac{z+1}{-8}
AI generated content. Review strictly for academic accuracy.