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If the direction rations of the normal of the plane which contains two parallel lines $\frac{x + 1}{3} = \frac{y - 2}{2} = \frac{z}{1} and \frac{x - 3}{3} = \frac{y + 4}{2} = \frac{z - 1}{1}$ are (a, 1, 13b) then select CORRECT alternative/s

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$a = 4 b$

$a + b = 6$

$a - 3 b = 2$

$A B - 16$

answer is A, C, D.

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Detailed Solution

Points on the first and second lines are (–1, 2, 0) and (3, –4, 1)

Equation of the plane is $8 (x + 1) + (y - 2) - 13 k (z) = 0$

$(3, - 4, 1)$ will also lie on it

$\Rightarrow 8 (4) - 6 - 13 K = 0 \Rightarrow 13 K = 26 \Rightarrow K = 0$

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