The plane containing the line frac{x - 3}{2} | vedclass

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(D) Let the given line be $L: \frac{x - 3}{2} = \frac{y + 2}{-1} = \frac{z - 1}{3} = k$. Any point on the line is $P(2k + 3, -k - 2, 3k + 1)$.The plan

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$(2, 0, -2)$

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(D) Let the given line be $L: \frac{x - 3}{2} = \frac{y + 2}{-1} = \frac{z - 1}{3} = k$. Any point on the line is $P(2k + 3, -k - 2, 3k + 1)$.The plane $P_1$ is $2x + 3y - z = 5$. The normal vector to $P_1$ is $\vec{n_1} = 2\hat{i} + 3\hat{j} - \hat{k}$.The direction of the line $L$ is $\vec{v} = 2\hat{i} - \hat{j} + 3\hat{k}$.The plane $P_2$ contains the line $L$ and its projection on $P_1$. This means $P_2$ contains the line $L$ and is perpendicular to $P_1$.The normal vector $\vec{n_2}$ of the required plane $P_2$ must be perpendicular to the direction of the line $\vec{v}$ and the normal of the plane $P_1$ $(\vec{n_1})$.$\vec{n_2} = \vec{v} 	imes \vec{n_1} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & -1 & 3 \ 2 & 3 & -1 \end{vmatrix} = \hat{i}(1 - 9) - \hat{j}(-2 - 6) + \hat{k}(6 + 2) = -8\hat{i} + 8\hat{j} + 8\hat{k}$.We can take the normal vector as $\vec{n} = \hat{i} - \hat{j} - \hat{k}$.The plane passes through the point $(3, -2, 1)$ on the line. The equation of the plane is $1(x - 3) - 1(y + 2) - 1(z - 1) = 0$.$x - 3 - y - 2 - z + 1 = 0 \Rightarrow x - y - z = 4$.Checking the options:$A: 2 - 2 - 0 = 0 
eq 4$$B: -2 - 2 - 2 = -6 
eq 4$$C: 0 - (-2) - 2 = 0 
eq 4$$D: 2 - 0 - (-2) = 4$. Thus,point $(2, 0, -2)$ lies on the plane.

The plane containing the line $\frac{x - 3}{2} = \frac{y + 2}{-1} = \frac{z - 1}{3}$ and also containing its projection on the plane $2x + 3y - z = 5$ contains which one of the following points?

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