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(A) 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 normal vector $\

Vedclass · Accepted Answer

$65$

Vedclass · Answer

(A) 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 normal vector $\vec{n} = \vec{AB} 	imes \vec{AC}$.$\vec{AB} = (1-1, 4-2, 1-0) = (0, 2, 1)$$\vec{AC} = (0-1, 5-2, 1-0) = (-1, 3, 1)$$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 0 & 2 & 1 \ -1 & 3 & 1 \end{vmatrix} = \hat{i}(2-3) - \hat{j}(0+1) + \hat{k}(0+2) = -\hat{i} - \hat{j} + 2\hat{k}$.The equation of the plane is $-1(x-1) - 1(y-2) + 2(z-0) = 0$,which simplifies to $x + y - 2z - 3 = 0$.Let the image of $P(1, 2, 6)$ be $Q(\alpha, \beta, \gamma)$. The formula for the image is $\frac{\alpha - x_0}{a} = \frac{\beta - y_0}{b} = \frac{\gamma - z_0}{c} = -2 \frac{ax_0 + by_0 + cz_0 + d}{a^2 + b^2 + c^2}$.Substituting the values: $\frac{\alpha - 1}{1} = \frac{\beta - 2}{1} = \frac{\gamma - 6}{-2} = -2 \frac{1(1) + 1(2) - 2(6) - 3}{1^2 + 1^2 + (-2)^2} = -2 \frac{1 + 2 - 12 - 3}{6} = -2 \frac{-12}{6} = 4$.Thus,$\alpha - 1 = 4 \Rightarrow \alpha = 5$,$\beta - 2 = 4 \Rightarrow \beta = 6$,and $\gamma - 6 = -8 \Rightarrow \gamma = -2$.Finally,$\alpha^2 + \beta^2 + \gamma^2 = 5^2 + 6^2 + (-2)^2 = 25 + 36 + 4 = 65$.

Let the image of the point $P(1, 2, 6)$ in the plane passing through the points $A(1, 2, 0)$,$B(1, 4, 1)$,and $C(0, 5, 1)$ be $Q(\alpha, \beta, \gamma)$. Then $(\alpha^2 + \beta^2 + \gamma^2)$ is equal to :

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