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 $P_{1} : 7 x + y + 2 z = 3, P_{2} : 3 x + 5 y - 6 z = 4.$ Let $a x + b y + c z = d$ the equation of the plane passing through the point of intersection of lines $L_{1}$ and $L_{2}$ and perpendicular to planes $P_{1}$ and $P_{2}$ .
Match List I with List II and select the correct answer using the code given below the lists.

$\begin{matrix} L i s t I & L i s t I I P & a = & 1 & 13 Q & b = & 2 & - 3 R & c = & 3 & 1 S & d = & 4 & - 2 \end{matrix}$

P-3, Q-2, R-4, S-1

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P-1, Q-3, R-4, S-2

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P-3, Q-2, R-1, S-4

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P-2, Q-4, R-1, S-3

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Solution

The correct option is A
P-3, Q-2, R-4, S-1

$L_{1} : \frac{x - 1}{2} = \frac{y - 0}{- 1} = \frac{z - (- 3)}{1}$
Normat of plane $P : n = ∣ ∣ ∣ ∣ ∣ \begin{matrix} ^i^j^k 712 35 - 6 \end{matrix} ∣ ∣ ∣ ∣ ∣ =^i (- 16) -^j (- 42 - 6) +^k (32) = - 16^i + 48^j + 32^k D R^{'} s o f n o r m a l n =^i - 3^j - 2^k P o i n t o f i n t e r s e c t i o n o f L_{1} a n d L_{2} . \Rightarrow 2 K_{1} + 1 = K_{2} + 4 a n d - k_{1} = k_{2} - 3 \Rightarrow k_{1} = 2 a n d k_{2} = 1 ∴ P o i n t o f i n t e r s e c t i o n (5, - 2, - 1) N o w e q u a t i o n o f p l a n e, 1. (x - 5) - 3 (y + 2) - 2 (z + 1) = 0 \Rightarrow x - 3 y - 2 z - 13 = 0 \Rightarrow x - 3 y - 2 z = 13 ∴ a \to 1, b \to - 3, c \to - 2, d \to 13$

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Similar questions

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 $P_{1} : 7 x + y + 2 z = 3, P_{2} : 3 x + 5 y - 6 z = 4.$ Let $a x + b y + c z = d$ the equation of the plane passing through the point of intersection of lines $L_{1}$ and $L_{2}$ and perpendicular to planes $P_{1}$ and $P_{2}$ .
Match List I with List II and select the correct answer using the code given below the lists.

$\begin{matrix} L i s t I & L i s t I I P & a = & 1 & 13 Q & b = & 2 & - 3 R & c = & 3 & 1 S & d = & 4 & - 2 \end{matrix}$

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 $P_{1}$ : $7 x + y + 2 z = 3, P_{2}$ : $3 x + 5 y - 6 z = 4.$ Let $a x + b y + c z = d$ be the equation of the plane passing through the point of intersection of lines $L_{1}$ and $L_{2}$ , and perpendicular to planes $P_{1}$ and $P_{2}$ .

Match List $1$ with List $2$ and select the correct answer using the code given below the lists:

Q. 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

P_{1} : 7 x + y + 2 z = 3,

P_{2} : 3 x + 5 y - 6 z = 4.

Let

a x + b y + c z = d

be the equation of plane passing through the point of intersection of the lines

L_{1}

and

L_{2}

, and perpendicular to the planes

P_{1}

and

P_{2}

.

Match

List - I

with the

List - II

and select the correct answer using the code given below the lists :

\begin{matrix} List I & List II P & a = & 1. & 13 Q & b = & 2. & - 3 R & c = & 3. & 1 S & d = & 4. & - 2 \end{matrix}

Match the four starting meterials (P, Q, R, S) given in List – I with the corresponding reaction schemes
(I, II, III, IV) provided in List – II and select the correct answer using the code given below the lists.

Q. List-II gives distance between pair of points given in List-I, match them correctly.

\begin{matrix} L i s t - I & L i s t I I (P) & (- 5, 7), (- 1, 3) & (1) & 5 (Q) & (5, 6), (1, 3) & (2) & \sqrt{8} (R) & (\sqrt{3} + 1, 1), (0, \sqrt{3}) & (3) & \sqrt{6} (S) & (0, 0), (- \sqrt{3}, \sqrt{3}) & (4) & 4 \sqrt{2} \end{matrix}

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