The sum of the intercepts on the coordinate a | vedclass

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(C) The equation of the 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 equat

Vedclass · Accepted Answer

$-4$

Vedclass · Answer

(C) The equation of the 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 points $(-2, -2, 2)$,$(1, -1, 2)$,and $(1, 1, 1)$:$\begin{vmatrix} x + 2 & y + 2 & z - 2 \ 1 - (-2) & -1 - (-2) & 2 - 2 \ 1 - (-2) & 1 - (-2) & 1 - 2 \end{vmatrix} = 0$$\Rightarrow \begin{vmatrix} x + 2 & y + 2 & z - 2 \ 3 & 1 & 0 \ 3 & 3 & -1 \end{vmatrix} = 0$Expanding the determinant:$(x + 2)(-1 - 0) - (y + 2)(-3 - 0) + (z - 2)(9 - 3) = 0$$-x - 2 + 3y + 6 + 6z - 12 = 0$$-x + 3y + 6z - 8 = 0$$x - 3y - 6z = -8$Dividing by $-8$ to get the intercept form $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$:$\frac{x}{-8} + \frac{y}{8/3} + \frac{z}{8/6} = 1$The intercepts are $a = -8$,$b = 8/3$,and $c = 4/3$.Sum of intercepts $= -8 + \frac{8}{3} + \frac{4}{3} = -8 + \frac{12}{3} = -8 + 4 = -4$.

The sum of the intercepts on the coordinate axes of the plane passing through the point $(-2, -2, 2)$ and containing the line joining the points $(1, -1, 2)$ and $(1, 1, 1)$ is

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