Let Q be the mirror image of the point P(1, 2 | vedclass

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(B) The mirror image $Q(x, y, z)$ of point $P(x_0, y_0, z_0)$ in the plane $ax + by + cz + d = 0$ is given by $\frac{x-x_0}{a} = \frac{y-y_0}{b} = \fr

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$(1, 2, 1)$

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(B) The mirror image $Q(x, y, z)$ of point $P(x_0, y_0, z_0)$ in the plane $ax + by + cz + d = 0$ is given by $\frac{x-x_0}{a} = \frac{y-y_0}{b} = \frac{z-z_0}{c} = -2 \frac{ax_0 + by_0 + cz_0 + d}{a^2 + b^2 + c^2}$.Substituting $P(1, 2, 1)$ and $x + 2y + 2z - 16 = 0$:$\frac{x-1}{1} = \frac{y-2}{2} = \frac{z-1}{2} = -2 \frac{1 + 2(2) + 2(1) - 16}{1^2 + 2^2 + 2^2} = -2 \frac{1 + 4 + 2 - 16}{9} = -2 \frac{-9}{9} = 2$.Thus,$x-1 = 2 \Rightarrow x = 3$,$y-2 = 4 \Rightarrow y = 6$,$z-1 = 4 \Rightarrow z = 5$. So $Q = (3, 6, 5)$.The plane $T$ passes through $Q(3, 6, 5)$ and contains the line passing through $A(0, 0, -1)$ with direction vector $\vec{v} = \hat{i} + \hat{j} + 2\hat{k}$.The vector $\vec{AQ} = (3-0)\hat{i} + (6-0)\hat{j} + (5 - (-1))\hat{k} = 3\hat{i} + 6\hat{j} + 6\hat{k}$.The normal to plane $T$ is $\vec{n} = \vec{AQ} 	imes \vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 3 & 6 & 6 \ 1 & 1 & 2 \end{vmatrix} = \hat{i}(12-6) - \hat{j}(6-6) + \hat{k}(3-6) = 6\hat{i} - 3\hat{k}$.Dividing by $3$,we use $\vec{n} = 2\hat{i} - \hat{k}$.The equation of plane $T$ is $2(x-0) + 0(y-0) - 1(z+1) = 0 \Rightarrow 2x - z = 1$.Checking the options: For $(1, 2, 1)$,$2(1) - 1 = 1$. Thus,$(1, 2, 1)$ lies on $T$.

Let $Q$ be the mirror image of the point $P(1, 2, 1)$ with respect to the plane $x + 2y + 2z = 16$. Let $T$ be a plane passing through the point $Q$ and containing the line $\vec{r} = -\hat{k} + \lambda(\hat{i} + \hat{j} + 2\hat{k}), \lambda \in R$. Then,which of the following points lies on $T$?

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