Show that the point (i-j+3k) and 3(i+j+k) are equidistant from the plane 5x+2y-7z+9=0 and lies on opposite side of it

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Please find below the solution to the asked query:

$$\begin{gathered} \mathrm{Given} \mathrm{points} \\ \mathrm{A}\left(\mathrm{i}-\mathrm{j}+3\mathrm{k}\right)=\left(1,-1,3\right) \\ \mathrm{B}\left(3\mathrm{i}+3\mathrm{j}+3\mathrm{k}\right)=\left(3,3,3\right) \\ \mathrm{Given} \mathrm{plane} = 5\mathrm{x}+2\mathrm{y}-7\mathrm{z}+9=0 \\ \mathrm{Perpendicular} \mathrm{distance} \mathrm{of} \mathrm{A} \mathrm{from} \mathrm{plane}=\left|\frac{5\times 1+2\times \left(-1\right)-7\times \left(3\right)+9}{\sqrt{25+4+49}}\right|=\left|\frac{-9}{\sqrt{78}}\right|=\frac{9}{\sqrt{78}}\mathrm{units} \\ \mathrm{Perpendicular} \mathrm{distance} \mathrm{of} \mathrm{B} \mathrm{from} \mathrm{plane}=\left|\frac{5\times 3+2\times 3-7\times 3+9}{\sqrt{25+4+49}}\right|=\left|\frac{9}{\sqrt{78}}\right|=\frac{9}{\sqrt{78}}\mathrm{units} \\ \mathrm{hence} \mathrm{both} \mathrm{points} \mathrm{are} \mathrm{equidistant} \mathrm{from} \mathrm{the} \mathrm{plane}. \\ \mathrm{Put} \mathrm{each} \mathrm{point} \mathrm{A} \mathrm{and} \mathrm{B} \mathrm{in} \mathrm{the} \mathrm{plane} \mathrm{equation} : \\ \mathrm{for} \mathrm{A}: \\ 5\times 1+2\times \left(-1\right)-7\times \left(3\right)+9=-9<0\left(-\mathrm{ve}\right) \\ \mathrm{for} \mathrm{B}: \\ 5\times 3+2\times 3-7\times 3+9=9 >0 \left(+\mathrm{ve}\right) \\ \mathbf{since}\mathbf{ }\mathbf{both}\mathbf{ }\mathbf{signs}\mathbf{ }\mathbf{are}\mathbf{ }\mathbf{opposite}\mathbf{ }\mathbf{to}\mathbf{ }\mathbf{each}\mathbf{ }\mathbf{other}\mathbf{ }\mathbf{the}\mathbf{ }\mathbf{points}\mathbf{ }\mathbf{lie}\mathbf{ }\mathbf{on}\mathbf{ }\mathbf{the}\mathbf{ }\mathbf{opposite}\mathbf{ }\mathbf{sides}\mathbf{.} \end{gathered}$$

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