Find the equation of a plane passing through the point P(6,5,9) and parallel to the plane determined by the points A(3,-1,2), B(5,2,4) and C(-1,-1,6). Also find the distance of this plane from the point A.

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Please find below the solution to the asked query:

$$\begin{gathered} \mathrm{Let} \mathrm{equation} \mathrm{of} \mathrm{the} \mathrm{required} \mathrm{plane} \mathrm{be} \\ \mathrm{ax}+\mathrm{by}+\mathrm{cz}+\mathrm{d}=0 \\ \mathrm{As} \mathrm{planes} \mathrm{are} \mathrm{parallel}, \mathrm{hence} \mathrm{their} \mathrm{direction} \mathrm{normal} \mathrm{will} \mathrm{be} \mathrm{same}. \\ \mathrm{Let} \mathrm{O} \mathrm{be} \mathrm{the} \mathrm{origin}. \\ \Rightarrow \overrightarrow{\mathrm{OA}}=3\hat{\mathrm{i}}-\hat{\mathrm{j}}+2\hat{\mathrm{k}} \\ \Rightarrow \overrightarrow{\mathrm{OB}}=5\hat{\mathrm{i}}+2\hat{\mathrm{j}}+4\hat{\mathrm{k}} \\ \Rightarrow \overrightarrow{\mathrm{OC}}=-\hat{\mathrm{i}}-\hat{\mathrm{j}}+6\hat{\mathrm{k}} \\ \Rightarrow \overrightarrow{\mathrm{AB}}=\overrightarrow{\mathrm{OB}}-\overrightarrow{\mathrm{OA}}=\left(5\hat{\mathrm{i}}+2\hat{\mathrm{j}}+4\hat{\mathrm{k}}\right)-\left(3\hat{\mathrm{i}}-\hat{\mathrm{j}}+2\hat{\mathrm{k}}\right)=2\hat{\mathrm{i}}+3\hat{\mathrm{j}}+2\hat{\mathrm{k}} \\ \Rightarrow \overrightarrow{\mathrm{AC}}=\overrightarrow{\mathrm{OC}}-\overrightarrow{\mathrm{OA}}=\left(-\hat{\mathrm{i}}-\hat{\mathrm{j}}+6\hat{\mathrm{k}}\right)-\left(3\hat{\mathrm{i}}-\hat{\mathrm{j}}+2\hat{\mathrm{k}}\right)=-2\hat{\mathrm{i}}+4\hat{\mathrm{k}} \\ \mathrm{Cross} \mathrm{product} \mathrm{of} \mathrm{above} \mathrm{two} \mathrm{vector} \mathrm{will} \mathrm{give} \mathrm{direction} \mathrm{normal}. \\ \Rightarrow \mathrm{a}\hat{\mathrm{i}}+\mathrm{b}\hat{\mathrm{j}}+\mathrm{c}\hat{\mathrm{k}}=\overrightarrow{\mathrm{AB}}\times \overrightarrow{\mathrm{AC}} \\ =\left|\begin{array}{ccc}\hat{\mathrm{i}}& \hat{\mathrm{j}}& \hat{\mathrm{k}}\\ 2& 3& 2\\ -2& 0& 4\end{array}\right|=\left(12-0\right)\hat{\mathrm{i}}-\left(8+4\right)\hat{\mathrm{j}}+\left(0+6\right)\hat{\mathrm{k}} \\ \Rightarrow \mathrm{a}\hat{\mathrm{i}}+\mathrm{b}\hat{\mathrm{j}}+\mathrm{c}\hat{\mathrm{k}}=12\hat{\mathrm{i}}-12\hat{\mathrm{j}}+6\hat{\mathrm{k}} \\ \mathrm{a}=12 \mathrm{and} \mathrm{b}=-12 \mathrm{and} \mathrm{c}=6 \\ \mathrm{Hence} \mathrm{equation} \mathrm{becomes}: \\ 12\mathrm{x}-12\mathrm{y}+6\mathrm{z}+\mathrm{d}=0 \\ \mathrm{Plane} \mathrm{passes} \mathrm{through} \left(6,5,9\right) \\ \Rightarrow 72-60+54+\mathrm{d}=0 \\ \Rightarrow \mathrm{d}=66 \\ \Rightarrow 12\mathrm{x}-12\mathrm{y}+6\mathrm{z}+66=0 \\ \Rightarrow \mathbf{2}\mathbf{x}\mathbf{-}\mathbf{2}\mathbf{y}\mathbf{+}\mathbf{z}\mathbf{+}\mathbf{11}\mathbf{=}\mathbf{0} \\ \mathrm{Distance} \mathrm{from} \mathrm{point} \mathrm{A}\left(3,-1,2\right) \mathrm{is} \mathrm{given} \mathrm{by}: \\ \mathrm{d}=\frac{\left|2\times 3-2\times \left(-1\right)+2+11\right|}{\sqrt{2^2+{\left(-2\right)}^2+1}}=\frac{\left|6+2+2+11\right|}{\sqrt{9}}=\frac{21}{3}=\mathbf{ }\mathbf{7}\mathbf{ }\mathbf{units} \end{gathered}$$
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