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(B) Let the direction ratios of the required line be $a, b, c$.Since the line is perpendicular to the lines with direction ratios $(-1, 2, 2)$ and $(0

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$2, -1, 2$

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(B) Let the direction ratios of the required line be $a, b, c$.Since the line is perpendicular to the lines with direction ratios $(-1, 2, 2)$ and $(0, 2, 1)$,we have the following equations:$-a + 2b + 2c = 0$ $(i)$$0a + 2b + c = 0$ (ii)From (ii),we get $c = -2b$.Substituting $c = -2b$ into $(i)$:$-a + 2b + 2(-2b) = 0$$-a + 2b - 4b = 0$$-a - 2b = 0 \implies a = -2b$.Now,we have $a = -2b$ and $c = -2b$.If we let $b = 1$,then $a = -2$ and $c = -2$.The direction ratios are $(-2, 1, -2)$,which is proportional to $(2, -1, 2)$.Alternatively,using the cross product of the two direction vectors $\vec{n_1} = -\hat{i} + 2\hat{j} + 2\hat{k}$ and $\vec{n_2} = 0\hat{i} + 2\hat{j} + 1\hat{k}$:$\vec{n_1} 	imes \vec{n_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ -1 & 2 & 2 \ 0 & 2 & 1 \end{vmatrix} = \hat{i}(2 - 4) - \hat{j}(-1 - 0) + \hat{k}(-2 - 0) = -2\hat{i} + 1\hat{j} - 2\hat{k}$.The direction ratios are proportional to $(-2, 1, -2)$,which is equivalent to $(2, -1, 2)$.

Direction ratios of the line which is perpendicular to the lines with direction ratios $-1, 2, 2$ and $0, 2, 1$ are

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