The direction cosines of the normal to the plane containing the lines having direction ratios 1, 2, 1 and 4, 5, –3 are

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$\frac{2}{\sqrt{5}}, \frac{- 1}{\sqrt{5}}, 0$

$\frac{- 11}{\sqrt{179}}, \frac{7}{\sqrt{179}}, \frac{- 3}{\sqrt{179}}$

$\frac{5}{\sqrt{41}}, \frac{- 4}{\sqrt{41}}, 0$

$\frac{1}{\sqrt{2}}, 0, \frac{- 1}{\sqrt{2}}$

answer is A.

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Detailed Solution

Let $b_{1} = \hat{i} + 2 \hat{j} + \hat{k}$ and $b_{2} = 4 \hat{i} + 5 \hat{j} - 3 \hat{k}$
Normal vector to plane
$b_{1} \times b_{2} = | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 1 \\ 4 & 5 & - 3 \end{matrix} | = \hat{i} (- 6 - 5) - \hat{j} (- 3 - 4) + \hat{k} (5 - 8) = - 11 \hat{i} + 7 \hat{j} - 3 \hat{k}$
$∴$ Direction cosines of normal to the plane
$= \frac{- 11}{\sqrt{{(- 11)}^{2} + 7^{2} + {(- 3)}^{2}}}, \frac{7}{\sqrt{{(- 11)}^{2} + 7^{2} + {(3)}^{2}}}, \frac{- 3}{\sqrt{{(- 11)}^{2} + 7^{2} + {(3)}^{2}}}$
$= \frac{- 11}{\sqrt{179}}, \frac{7}{\sqrt{179}}, \frac{- 3}{\sqrt{179}}$