Find the equation of the plane that passes th | vedclass

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(A) Let the given points be $A(1, 1, -1)$,$B(6, 4, -5)$,and $C(-4, -2, 3)$.To check if these points are collinear,we calculate the determinant of the

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The points are collinear,so there are infinitely many planes.

Vedclass · Answer

(A) Let the given points be $A(1, 1, -1)$,$B(6, 4, -5)$,and $C(-4, -2, 3)$.To check if these points are collinear,we calculate the determinant of the matrix formed by the vectors:$\begin{vmatrix} 1 & 1 & -1 \ 6 & 4 & -5 \ -4 & -2 & 3 \end{vmatrix} = 1(12 - 10) - 1(18 - 20) + (-1)(-12 + 16)$$= 1(2) - 1(-2) - 1(4)$$= 2 + 2 - 4 = 0$Since the determinant is $0$,the points $A$,$B$,and $C$ are collinear.Because the points are collinear,they do not uniquely define a plane. Therefore,there are infinitely many planes passing through these three points.

Find the equation of the plane that passes through the three points $(1, 1, -1)$,$(6, 4, -5)$,and $(-4, -2, 3)$.

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