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Question

(A) The equation of a plane passing through three points $(x_1, y_1, z_1)$,$(x_2, y_2, z_2)$,and $(x_3, y_3, z_3)$ is given by the determinant equatio

Vedclass · Accepted Answer

$5x + 6y + 2z - 23 = 0$

Vedclass · Answer

(A) The equation of a plane passing through three points $(x_1, y_1, z_1)$,$(x_2, y_2, z_2)$,and $(x_3, y_3, z_3)$ is given by the determinant equation: $\begin{vmatrix} x-x_1 & y-y_1 & z-z_1 \ x_2-x_1 & y_2-y_1 & z_2-z_1 \ x_3-x_1 & y_3-y_1 & z_3-z_1 \end{vmatrix} = 0$ Substituting the given points $(1, 2, 3)$,$(-1, 4, 2)$,and $(3, 1, 1)$: $\begin{vmatrix} x-1 & y-2 & z-3 \ -1-1 & 4-2 & 2-3 \ 3-1 & 1-2 & 1-3 \end{vmatrix} = 0$ $\begin{vmatrix} x-1 & y-2 & z-3 \ -2 & 2 & -1 \ 2 & -1 & -2 \end{vmatrix} = 0$ Expanding along the first row: $(x-1)[(2)(-2) - (-1)(-1)] - (y-2)[(-2)(-2) - (2)(-1)] + (z-3)[(-2)(-1) - (2)(2)] = 0$ $(x-1)[-4 - 1] - (y-2)[4 + 2] + (z-3)[2 - 4] = 0$ $-5(x-1) - 6(y-2) - 2(z-3) = 0$ $-5x + 5 - 6y + 12 - 2z + 6 = 0$ $-5x - 6y - 2z + 23 = 0$ Multiplying by $-1$,we get: $5x + 6y + 2z - 23 = 0$

The equation of the plane passing through the points $(1, 2, 3)$,$(-1, 4, 2)$ and $(3, 1, 1)$ is

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