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Question

Find the equation of the plane containing the points $$(2, 3,-1), (4, 5, 2), (3, 6, 5)$$


Solution

Now,
$$\begin{array}{l} a\left( { x-2 } \right) +b\left( { y-3 } \right) +c\left( { z+1 } \right) =0 \\ 2a+2b+3c=0 \\ a+3b+6c=0 \\ \frac { a }{ { 12-9 } } =\frac { { -b } }{ { 12-3 } } =\frac { c }{ { 6-2 } }  \\ \frac { a }{ 3 } =\frac { b }{ { -9 } } =\frac { c }{ 4 }  \\ \Rightarrow 3\left( { x-2 } \right) -9\left( { y-3 } \right) +4\left( { z+1 } \right) =0 \\ 3x-9y+4z+25=0 \end{array}$$

Mathematics

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