If the direction ratios of the normal of the plane which contains two parallel lines x+13=y−22=z1 and x−33=y+42 =z−11 are (a,1,13b) then

i^ j^ k^3 2 14 −6 1 =8i^+j^−26k^=(8,1,−26)∴a=8 ,13b=−26 b=−2

If the direction ratios of the normal of the plane which contains two parallel lines x+13=y−22=z1 and x−33=y+42 =z−11 are (a,1,13b) then

i^ j^ k^3 2 14 −6 1 =8i^+j^−26k^=(8,1,−26)∴a=8 ,13b=−26 b=−2

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If the direction ratios of the normal of the plane which contains two parallel lines $\frac{x + 1}{3} = \frac{y - 2}{2} = \frac{z}{1} a n d \frac{x - 3}{3} = \frac{y + 4}{2} = \frac{z - 1}{1}$ are $(a, 1, 13 b)$ then

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a=4b

a+b=6

a+3b=2

ab=16

answer is B, C.

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$\begin{array}{l} |\begin{array}{l} \hat{i} \hat{j} \hat{k} \\ 3 2 1 \\ 4 - 6 1 \end{array}| = 8 \hat{i} + \hat{j} - 26 \hat{k} = (8, 1, - 26) \\ ∴ a = 8, 13 b = - 26 b = - 2 \end{array}$

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If the direction ratios of the normal of the plane which contains two parallel lines x+13=y−22=z1 and x−33=y+42 =z−11 are (a,1,13b) then

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If the direction ratios of the normal of the plane which contains two parallel lines $\frac{x + 1}{3} = \frac{y - 2}{2} = \frac{z}{1} a n d \frac{x - 3}{3} = \frac{y + 4}{2} = \frac{z - 1}{1}$ are $(a, 1, 13 b)$ then