Consider the lines and the planes P1 : 7x + y + 2z = 3, P2 : 3x + 5y – 6z = 4. Let ax + by + cz = d be the equation of the plane passing through the point of intersection of lines L1 and L2, and perpendicular to planes P1 and P2.Match List-I with List-II and select the correct answer using the code given below the lists:List IList II(P) a =(1) 13(Q) b =(2) – 3(R) c =(3) 1(S) d =(4)

Engineering

Mathematics

Equation of Straight Line in 3D

Question

Consider the lines $L_{1} : \frac{x - 1}{2} = \frac{y}{- 1} = \frac{z + 3}{1}, L_{2} : \frac{x - 4}{1} = \frac{y + 3}{1} = \frac{z + 3}{2}$ and the planes P₁ : 7x + y + 2z = 3, P₂ : 3x + 5y – 6z = 4. Let ax + by + cz = d be the equation of the plane passing through the point of intersection of lines L₁ and L₂, and perpendicular to planes P₁ and P₂.

Match List-I with List-II and select the correct answer using the code given below the lists:

List I	List II
(P) a =	(1) 13
(Q) b =	(2) – 3
(R) c =	(3) 1
(S) d =	(4) – 2

P → 3

Q → 2

R → 4

S → 1

P → 2

Q → 4

R → 1

S → 3

P → 3

Q → 2

R → 1

S → 4

P → 1

Q → 3

R → 4

S → 2

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Any poin on L₁ is (2λ + 1, – λ, λ – 3) and any point on L₂ is (µ + 4, µ – 3, 2µ – 3)

$∴$ λ = 2, µ = 1

So, point of intersection is (5, – 2, – 1).

Also, normal vector of required plane is parallel to vector = $| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 7 & 1 & 2 \\ 3 & 5 & - 6 \end{matrix} | = - 16 (\hat{i} - 3 \hat{j} - 2 \hat{k})$

So, equation of required plane is 1 (x – 5) – 3 (y + 2) – 2 (z + 1) = 0

⇒ x – 5y – 2z = 13

$∴$ a = 1, b = – 3, c = – 2, d = 13.