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(B) The plane passes through the point $\vec{a} = 3 \hat{i} + 2 \hat{j} + 6 \hat{k}$ and is parallel to the vectors $\vec{b} = 2 \hat{i} + \hat{j} + \

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$2 x-y-3 z=-14$

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(B) The plane passes through the point $\vec{a} = 3 \hat{i} + 2 \hat{j} + 6 \hat{k}$ and is parallel to the vectors $\vec{b} = 2 \hat{i} + \hat{j} + \hat{k}$ and $\vec{c} = \hat{i} - \hat{j} + \hat{k}$.The normal vector $\vec{N}$ to the plane is given by the cross product of the two parallel vectors:$\vec{N} = \vec{b} 	imes \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 1 & 1 \ 1 & -1 & 1 \end{vmatrix}$$\vec{N} = \hat{i}(1 - (-1)) - \hat{j}(2 - 1) + \hat{k}(-2 - 1) = 2 \hat{i} - \hat{j} - 3 \hat{k}$.The equation of the plane is $(\vec{r} - \vec{a}) \cdot \vec{N} = 0$,where $\vec{r} = x \hat{i} + y \hat{j} + z \hat{k}$.$((x-3) \hat{i} + (y-2) \hat{j} + (z-6) \hat{k}) \cdot (2 \hat{i} - \hat{j} - 3 \hat{k}) = 0$$2(x-3) - 1(y-2) - 3(z-6) = 0$$2x - 6 - y + 2 - 3z + 18 = 0$$2x - y - 3z + 14 = 0$$2x - y - 3z = -14$.

The equation of the plane passing through $3 \hat{i}+2 \hat{j}+6 \hat{k}$ and parallel to the vectors $2 \hat{i}+\hat{j}+\hat{k}$ and $\hat{i}-\hat{j}+\hat{k}$ is

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