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Find the vector equation of the plane through the point A(3,-2,1)\, and perpendicular to the vector 4{\bar  {i}}+7{\bar  {j}}-4{\bar  {k}}\,.

Given the required plane passes through the point A(3,-2,1)=3{\bar  {i}}-2{\bar  {j}}+{\bar  {k}}\,

Also given vector is perpendicular to 4{\bar  {i}}+7{\bar  {j}}-4{\bar  {k}}\,.

Therefore the unit vector perpendicular to the plane is {\frac  {4{\bar  {i}}+7{\bar  {j}}-4{\bar  {k}}}{{\sqrt  {16+49+16}}}}={\frac  {1}{9}}(4{\bar  {i}}+7{\bar  {j}}-4{\bar  {k}})\,.

Therefore,the vector equation of the plane is ({\bar  {r}}-{\bar  {a}})\cdot {\bar  {n}}=0,{\bar  {r}}\cdot {\bar  {n}}={\bar  {a}}\cdot {\bar  {n}}\,

Hence {\bar  {r}}\cdot {\frac  {1}{9}}(4{\bar  {i}}+7{\bar  {j}}-4{\bar  {k}})=(3{\bar  {i}}-2{\bar  {j}}+{\bar  {k}})\cdot {\frac  {1}{9}}(4{\bar  {i}}+7{\bar  {j}}-4{\bar  {k}})\,.

{\bar  {r}}\cdot (4{\bar  {i}}+7{\bar  {j}}-4{\bar  {k}})=12-14-4=-6\,.

{\bar  {r}}\cdot (-4{\bar  {i}}-7{\bar  {j}}+4{\bar  {k}})=6\,

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