(a) Find the equation of the plane passing through points (2, 1, 0), (3, -2, -2) and (1, 1, -7). Also, obtain its distance from the origin.

(a)Equation of plane is given by\[\begin{vmatrix} x-2 & y-1 & z \\ 1 & -3 & -2 \\ -1 & 0 & -7 \end{vmatrix} = 0\]\[21(x-2) + 9(y-1) - 3z = 0\]i.e., $\displaystyle 7x + 3y - z = 17$Distance of plane from origin is\[d = \frac{|0+0-0-17|}{\sqrt{59}} = \frac{17}{\sqrt{59}}\]Or(b)For lines $\displaystyle \dfrac{x}{1} = \dfrac{y-1}{2} = \dfrac{z-2}{3}$ and $\displaystyle \dfrac{x+1}{1} = \dfrac{y+2}{2} = \dfrac{z-1}{3}$Let $\displaystyle \vec{a_1} = \hat{j} + 2\hat{k},\ \vec{b} = \hat{i} + 2\hat{j} + 3\hat{k}$$\displaystyle \vec{a_2} = -\hat{i} - 2\hat{j} + \hat{k},\ \vec{b} = \hat{i} + 2\hat{j} + 3\hat{k}$Clearly lines are parallelHence, Shortest distance or distance is given by\[\frac{\left|(\vec{a_2} - \vec{a_1}) \times \vec{b}\right|}{|\vec{b}|}\]\[\vec{a_2} - \vec{a_1} = -\hat{i} - 3\hat{j} - \hat{k}\]\[(\vec{a_2} - \vec{a_1}) \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & -3 & -1 \\ 1 & 2 & 3 \end{vmatrix}\]\[= -7\hat{i} + 2\hat{j} + \hat{k}\]Required distance $\displaystyle = \dfrac{\sqrt{49+4+1}}{\sqrt{1+4+9}} = \dfrac{\sqrt{27}}{\sqrt{7}}$ or $\displaystyle \dfrac{3\sqrt{21}}{7}$