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The equation of the line $l$ is given by $\vec{r}=4\hat{i}+2\hat{j}+2\hat{k}+\lambda(\hat{i}-\hat{j}-\hat{k})$. A general point P on this line can be represented as $P(4+\lambda, 2-\lambda, 2-\lambda)$.
The coordinates of point A are (2, 1, 2). The direction ratios of the line segment AP are given by $(4+\lambda - 2, 2-\lambda - 1, 2-\lambda - 2)$, which simplifies to $(\lambda+2, 1-\lambda, -\lambda)$.
Since AP is perpendicular to the line $l$, the dot product of their direction ratios must be zero. The direction ratios of line $l$ are (1, -1, -1). Therefore,\r\n$(\lambda+2)(1) + (1-\lambda)(-1) + (-\lambda)(-1) = 0$\r\n$\lambda + 2 - 1 + \lambda + \lambda = 0$\r\n$3\lambda + 1 = 0$\r\n$\lambda = -\frac{1}{3}$
Substitute $\lambda = -\frac{1}{3}$ into the coordinates of point P: $P(4-\frac{1}{3}, 2+\frac{1}{3}, 2+\frac{1}{3}) = P(\frac{11}{3}, \frac{7}{3}, \frac{7}{3})$.
Since P is the midpoint of AA', let A' be $(x, y, z)$. Then,\r\n$\frac{x+2}{2} = \frac{11}{3}$, $\frac{y+1}{2} = \frac{7}{3}$, $\frac{z+2}{2} = \frac{7}{3}$\r\n$x+2 = \frac{22}{3}$, $y+1 = \frac{14}{3}$, $z+2 = \frac{14}{3}$\r\n$x = \frac{22}{3} - 2 = \frac{16}{3}$, $y = \frac{14}{3} - 1 = \frac{11}{3}$, $z = \frac{14}{3} - 2 = \frac{8}{3}$\r\nSo, A' is $(\frac{16}{3}, \frac{11}{3}, \frac{8}{3})$.
The direction ratios of the line AA' are $(\frac{16}{3} - 2, \frac{11}{3} - 1, \frac{8}{3} - 2) = (\frac{10}{3}, \frac{8}{3}, \frac{2}{3})$. We can use the direction ratios (5, 4, 1).\r\nThe equation of the line AA' passing through A(2, 1, 2) is given by:\r\n$\frac{x-2}{5} = \frac{y-1}{4} = \frac{z-2}{1}$
\r\n Final Answer: Image A' = $(\frac{16}{3}, \frac{11}{3}, \frac{8}{3})$, Equation of line AA': $\frac{x-2}{5} = \frac{y-1}{4} = \frac{z-2}{1}$, Foot of perpendicular P = $(\frac{11}{3}, \frac{7}{3}, \frac{7}{3})$<\/span>\r\n <\/p>\r\n <\/div>\r\n <\/div>