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

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$\hat{i}+\hat{j}-\hat{k}$

Vedclass · Answer

(C) The equation of a plane passing through three points $A(x_1, y_1, z_1)$, $B(x_2, y_2, z_2)$, and $C(x_3, y_3, z_3)$ is given by the determinant equation: $\begin{vmatrix} x-x_1 & y-y_1 & z-z_1 \ x_2-x_1 & y_2-y_1 & z_2-z_1 \ x_3-x_1 & y_3-y_1 & z_3-z_1 \end{vmatrix} = 0$ Given points are $A(1, -1, 1)$, $B(1, -2, 3)$, and $C(1, 2, -3)$. Substituting these into the determinant: $\begin{vmatrix} x-1 & y+1 & z-1 \ 1-1 & -2-(-1) & 3-1 \ 1-1 & 2-(-1) & -3-1 \end{vmatrix} = 0$ $\begin{vmatrix} x-1 & y+1 & z-1 \ 0 & -1 & 2 \ 0 & 3 & -4 \end{vmatrix} = 0$ Expanding along the first column: $(x-1)((-1)(-4) - (2)(3)) - 0 + 0 = 0$ $(x-1)(4-6) = 0$ $-2(x-1) = 0$ $x-1 = 0$ $x = 1$ Now, check the options to see which point satisfies $x=1$: Option $C$ is $\hat{i}+\hat{j}-\hat{k}$, which corresponds to the point $(1, 1, -1)$. Since $x=1$ for this point, it lies on the plane.

$A$ point lying on the plane that passes through the points $\hat{i}-\hat{j}+\hat{k}$, $\hat{i}-2\hat{j}+3\hat{k}$, and $\hat{i}+2\hat{j}-3\hat{k}$ is:

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