The plane passing through the point (4, -1, 2 | vedclass

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(B) The normal vector $\vec{n}$ to the plane is the cross product of the direction vectors of the two lines,$\vec{n_1} = (3, -1, 2)$ and $\vec{n_2} =

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$(1, 1, 1)$

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(B) The normal vector $\vec{n}$ to the plane is the cross product of the direction vectors of the two lines,$\vec{n_1} = (3, -1, 2)$ and $\vec{n_2} = (1, 2, 3)$.$\vec{n} = \vec{n_1} 	imes \vec{n_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 3 & -1 & 2 \ 1 & 2 & 3 \end{vmatrix} = \hat{i}(-3 - 4) - \hat{j}(9 - 2) + \hat{k}(6 + 1) = -7\hat{i} - 7\hat{j} + 7\hat{k}$.We can take the normal vector as $\vec{n} = (1, 1, -1)$ by dividing by $-7$.The equation of the plane passing through $(4, -1, 2)$ with normal $(1, 1, -1)$ is $1(x - 4) + 1(y + 1) - 1(z - 2) = 0$.$x - 4 + y + 1 - z + 2 = 0 \Rightarrow x + y - z = 1$.Checking the options:For $(1, 1, 1)$,$1 + 1 - 1 = 1$. This satisfies the equation.

The plane passing through the point $(4, -1, 2)$ and parallel to the lines $\frac{x + 2}{3} = \frac{y - 2}{-1} = \frac{z + 1}{2}$ and $\frac{x - 2}{1} = \frac{y - 3}{2} = \frac{z - 4}{3}$ also passes through the point

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