A plane passing through the point (3,1,1) con | vedclass

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(B) The normal vector $\overrightarrow{n}$ to the plane is perpendicular to both lines. Thus,$\overrightarrow{n}$ is the cross product of the directio

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$5$

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(B) The normal vector $\overrightarrow{n}$ to the plane is perpendicular to both lines. Thus,$\overrightarrow{n}$ is the cross product of the direction vectors of the two lines:$\overrightarrow{n} = (\hat{i} - 2\hat{j} + 2\hat{k}) 	imes (2\hat{i} + 3\hat{j} - \hat{k})$$\overrightarrow{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & -2 & 2 \ 2 & 3 & -1 \end{vmatrix} = \hat{i}(2 - 6) - \hat{j}(-1 - 4) + \hat{k}(3 + 4) = -4\hat{i} + 5\hat{j} + 7\hat{k}$The equation of the plane passing through $(3, 1, 1)$ with normal vector $\overrightarrow{n} = -4\hat{i} + 5\hat{j} + 7\hat{k}$ is:$-4(x - 3) + 5(y - 1) + 7(z - 1) = 0$$-4x + 12 + 5y - 5 + 7z - 7 = 0$$-4x + 5y + 7z = 0$Since the plane passes through $(\alpha, -3, 5)$,we substitute these coordinates into the plane equation:$-4(\alpha) + 5(-3) + 7(5) = 0$$-4\alpha - 15 + 35 = 0$$-4\alpha + 20 = 0$$4\alpha = 20$$\alpha = 5$

$A$ plane passing through the point $(3,1,1)$ contains two lines whose direction ratios are $1, -2, 2$ and $2, 3, -1$ respectively. If this plane also passes through the point $(\alpha, -3, 5)$,then $\alpha$ is equal to

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