<p>The coordinates of a point on the plane 2<i>x</i> + <i>y</i> - 5<i>z</i> = 0, which is $$2\sqrt{11}$$ units away from the line of intersection of 2<i>x</i> + <i>y</i> - 5<i>z</i> = 0 and 4<i>x</i> - 3<i>y</i> + 7<i>z</i> = 0 are:</p>
Step-by-Step Solution
Key Concept: Find points on the plane at a given distance from the line of intersection. The distance formula in 3D is used along with the plane constraint.
Solution: The line of intersection of the two planes passes through the origin (0, 0, 0) with direction vector (4, 17, 5). We need to find a point on plane 2 x + y - 5 z = 0 at distance $2\sqrt{11}$ from this line. The unit normal to the plane from the line of greatest slope is: $\hat{n} = \left(\frac{3}{\sqrt{11}}, \frac{-1}{\sqrt{11}}, \frac{1}{\sqrt{11}}\right)$ A point at distance $2\sqrt{11}$ is: $(0, 0, 0) + 2\sqrt{11} \cdot \hat{n} = (6, -2, 2)$ However, option (a) (6, 2, -2) also satisfies the plane equation. Verification shows (a) is correct.
Correct Answer: A