Question

Let planes P₁ : 3x + 4y + (k² – 4)z = λ and P₂ : –3x – ky + 6 = 0, where k, λ $\in$ R intersect the y-axis at a point whose distance from the origin is 3 units. The points (0, –4, 0) and (0, 0, 0) lie on opposite sides of both the planes.

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Linked Question 1

If the plane perpendicular to P₁ and parallel to z-axis and containing the point $(\frac{α}{2}, \frac{2 - α^{2}}{3}, 0), α \in R$ for exactly one real value of α is 8x + by + cz + d = 0, where b, c, d $\in$ I then |b + c + d| equals

(ii) Plane perpendicular to the plane P₁ and parallel to z-axis is 4x – 3y + q = 0

$(\frac{α}{2}, \frac{2 - α^{2}}{3}, 0)$ lies on the plane

2α – (2 – α²) + q = 0

α² + 2α + q – 2 = 0

For exactly one real value of α

D = 0

⇒ 4 – 4 (q – 2) = 0 ⇒ q = 3

$∴$ 4x – 3y + 3 = 0

8x – 6y + 6 = 0 $\equiv$ 8x + by + cz + d = 0

$∴$ b + c + d = 0.

Linked Question 2

The equation of the plane containing line of intersection of P₁ and P₂ and also a point whose position vector $2 \hat{i} + 3 \hat{j} + \hat{k}$ is

3x – 2y = 6

3x – y – 3 = 0

3x – y – 6 = 0

2x – 3y = 9

Putting x = z = 0 and y = 3 in the planes.

k = 2 and λ = 12

P₁ : 3x + 4y – 12 = 0

and P₂ : –3x – 2y + 6 = 0

But points (0, 0, 0) and (0, –4, 0)

lie on opposite sides of both the plane.

$∴$ Putting x = 0 = z = 0 and y = –3

in the planes, k = –2 and λ = –12

$∴$ P₁ : 3x + 4y = –12 and

P₂ : –3x + 2y + 6 = 0

(i) Required plane is P₁ + tP₂ = 0

3x + 4y + 12 + t (– 3x + 2y + 16) = 0

Putting (2, 3, 1), get t = – 5

$∴$ 18x – 6y – 18 = 0

⇒ 3x – y – 3 = 0.