1. **Problem statement:** Find the image $A'$ of the point $A(2,1,2)$ in the line given by the parametric form $\mathbf{r} = \mathbf{a} + \lambda \mathbf{d}$ where $\mathbf{a} = (1,7,4)$ and $\mathbf{d} = (1,-1,-1)$. Also find the equation of the line joining $A$ and $A'$, and find the foot of the perpendicular from $A$ to the line $l$. 2. **Understanding the problem:** The image $A'$ of point $A$ in the line $l$ is the reflection of $A$ about $l$. To find $A'$, we first find the foot of the perpendicular $F$ from $A$ to $l$, then use the midpoint formula for reflection: $F$ is the midpoint of $AA'$, so $\mathbf{A'} = 2\mathbf{F} - \mathbf{A}$. 3. **Step 1: Find the foot of perpendicular $F$ from $A$ to line $l$.** The line $l$ is given by: $$\mathbf{r} = (1,7,4) + \lambda (1,-1,-1)$$ Let the foot of perpendicular be $F = (x,y,z) = (1 + \lambda, 7 - \lambda, 4 - \lambda)$. The vector $\overrightarrow{AF} = (1 + \lambda - 2, 7 - \lambda - 1, 4 - \lambda - 2) = (\lambda -1, 6 - \lambda, 2 - \lambda)$. Since $AF$ is perpendicular to the direction vector $\mathbf{d} = (1,-1,-1)$, their dot product is zero: $$\overrightarrow{AF} \cdot \mathbf{d} = 0$$ $$ (\lambda -1)(1) + (6 - \lambda)(-1) + (2 - \lambda)(-1) = 0 $$ $$ \lambda -1 -6 + \lambda -2 + \lambda = 0 $$ $$ 3\lambda -9 = 0 $$ $$ 3\lambda = 9 $$ $$ \lambda = 3 $$ 4. **Calculate coordinates of $F$:** $$ F = (1 + 3, 7 - 3, 4 - 3) = (4,4,1) $$ 5. **Find the image point $A'$:** Using midpoint formula for reflection: $$ F = \frac{A + A'}{2} \implies A' = 2F - A $$ $$ A' = 2(4,4,1) - (2,1,2) = (8,8,2) - (2,1,2) = (6,7,0) $$ 6. **Equation of line joining $A$ and $A'$:** Direction vector: $$ \overrightarrow{AA'} = (6-2, 7-1, 0-2) = (4,6,-2) $$ Parametric form: $$ \mathbf{r} = (2,1,2) + \mu (4,6,-2) $$ **Final answers:** - Image point $A' = (6,7,0)$ - Equation of line $AA'$: $$\mathbf{r} = (2,1,2) + \mu (4,6,-2)$$ - Foot of perpendicular $F = (4,4,1)$