Question 6

I need a plane containing the line through (1,3,-2) with direction (2,3,1), perpendicular to the plane x+2y+3z=4. Taking the cross product of (2,3,1) and (1,2,3) gives the normal (7,−5,1), so I'm computing the plane equation using the point (1,3,-2) with this normal. Checking option A confirms the plane equation is 7x−5y+z=−10, which is true. For B, I compute a parallel plane through (4,2,2) and find the distance between them is 2√3, so that's false. For C, the distance from the origin to plane P works out to 2/√3, also false. For D, comparing the normal with (2,2,1) using the dot product gives cos θ = 1/(3√3), matching the claim, so D is true. **Normal of P:** n = (2,3,1) × (1,2,3) = (7, −5, 1). Through (1,3,−2): 7 − 15 − 2 = −10 → P: 7x − 5y + z = −10. (A) true. **P₁** through (4,2,2): 7x − 5y + z = 20. Distance between P and P₁ = 30/√75 = 2√3, not 30. (B) false. **Distance of P from origin** = 10/√75 = 2/√3 ≠ 2√3. (C) false. **Angle with 2x+2y+z=3:** cos θ = |14 − 10 + 1|/(√75 · 3) = 5/(15√3) = 1/(3√3). (D) true.

No written reasoning was returned for this item.

**Clarifying final answer format** I’m noticing that the user wants a final answer in JSON format only, with no additional commentary afterward. The developer's instructions state reasoning can precede the final answer, but the last characters must be a single JSON object. I’m thinking it’s best to ensure the conclusion is structured appropriately, just as it mentions ending with JSON only. It seems like clarity is key here, and I should focus on that.

I need a plane containing the given line, so I take the cross product of the line's direction with the normal of the given plane to find a normal vector for plane P, getting (7,-5,1), and I'm setting up the plane equation through the point (1,3,-2). 2)=0 => 7x-5y+z+10=0, matching option A, and I verify the point satisfies this. Then I compute the distance from the origin, getting 2/√3, which rules out option C. Next I build plane P1 through (4,2,2) parallel to P, getting 7x-5y+z=20, then find the distance between P and P1 as 2√3—so option B's claim of distance 30 is false. Finally I start computing the angle between P and the plane 2x+2y+z=3 for option D, taking the dot product of normals (7,-5,1) and (2,2,1). For D, this gives cos θ = 5/(15√3) = 1/(3√3), confirming D is correct. So the correct answers are A and D. Looking at this problem, I need to find plane P containing the given line and perpendicular to x+2y+3z=4. **Finding Plane P:** - Line direction vector: (2, 3, 1), passing through (1, 3, -2) - Normal to given plane: (1, 2, 3) - Normal to P = (2,3,1) × (1,2,3) = (9-2, 1-6, 4-3) = (7, -5, 1) Plane P: 7(x-1) - 5(y-3) + 1(z+2) = 0 → **7x - 5y + z = -10** Verify: 7(1) - 5(3) + (-2) = 7-15-2 = -10 ✓ **(A) is TRUE** **Distance from origin to P:** Distance = |10|/√(49+25+1) = 10/√75 = 10/(5√3) = 2/√3 ≈ 1.155 This is **not** 2√3, so **(C) is FALSE** **Finding P1 and distance between P and P1:** P1 passes through (4,2,2), parallel to P (same normal (7,-5,1)): 7(x-4) - 5(y-2) + 1(z-2) = 0 → 7x - 5y + z = 20 Distance between planes: |20 - (-10)|/√75 = 30/(5√3) = 6/√3 = 2√3 The distance is **2√3, not 30**, so **(B) is FALSE** **Angle between P and 2x+2y+z=3:** Normal of P: (7,-5,1), Normal of given plane: (2,2,1) cos θ = |7(2) + (-5)(2) + 1(1)| / (√75 · √9) = |14-10+1| / (5√3 · 3) = 5/(15√3) = 1/(3√3) This matches exactly, so **(D) is TRUE** **Conclusion:** Statements A and D are true. ```json