Let the plane $P: 8x + \alpha_1 y + \alpha_2 z + 12 = 0$ be parallel to the line $L: \frac{x+2}{2} = \frac{y-3}{3} = \frac{z+4}{5}$. If the intercept of P on the y-axis is 1, then the distance between P and L is:
Step-by-Step Solution
Key Concept: Parallel plane condition: $\vec{n}\cdot\vec{d}=0$. Y-intercept gives another equation. Then compute distance from a point on L to plane P.
$\alpha_1=-12, \alpha_2=4$. Plane: $8x-12y+4z+12=0$, i.e., $2x-3y+z+3=0$. Distance from $(-2,3,-4)$: $\frac{|-4-9-4+3|}{\sqrt{14}} = \frac{14}{\sqrt{14}} = \sqrt{14}$. Answer: (1)
Correct Answer: $\sqrt{14}$