### Instruction: Let $\ell$ be the line in space through the points $(0,0,1)$ and $(1,1,1).$  Let $d$ be the length of the shortest possible path that begins at the origin, travels to a point on $\ell,$ then ends at the point $(1,0,0).$  Enter $d^2.$.### Response: To visualize the problem, I draw a sketch of the line $\ell$ and the points involved.
I label the point on $\ell$ that I want to find as $P.$
The shortest possible path from the origin to $P$ is the line segment $\overline{OP},$ which is perpendicular to $\ell.$
Similarly, the shortest possible path from $P$ to $(1,0,0)$ is the line segment $\overline{PQ},$ where $Q = (1,0,0).$
So the length of the path I want is $d = OP + PQ.$
To find $OP,$ I need to find the coordinates of $P.$
To do that, I need to find the equation of the line $\ell.$
Since $\ell$ passes through two given points, I can use the parametric form of the line equation: $x = x_0 + at, y = y_0 + bt, z = z_0 + ct,$ where $(x_0, y_0, z_0)$ is a point on the line and $(a, b, c)$ is a direction vector.
Choosing $(0,0,1)$ as the point on the line, I get $x_0 = 0, y_0 = 0, z_0 = 1.$
To find the direction vector, I subtract the coordinates of the two points: $(a, b, c) = (1,1,1) - (0,0,1) = (1,1,0).$
So the equation of the line $\ell$ is $x = t, y = t, z = 1.$
Now I need to find the value of $t$ that makes the line segment $\overline{OP}$ perpendicular to $\ell.$
One way to do that is to use the dot product of the vectors $\overrightarrow{OP}$ and $(a, b, c),$ which should be zero if they are perpendicular.
The vector $\overrightarrow{OP}$ has components $(x - 0, y - 0, z - 0) = (x, y, z),$ so the dot product is $(x, y, z) \cdot (1, 1, 0) = x + y.$
Substituting the equation of the line $\ell$ into the dot product, I get $x + y = t + t = 2t.$
So I need to solve $2t = 0$ for $t,$ which gives me $t = 0.$
Plugging $t = 0$ into the equation of the line $\ell,$ I get the coordinates of $P:$ $x = 0, y = 0, z = 1.$
So $P = (0,0,1).$
To find $OP,$ I use the distance formula between the origin and $P:$ $OP = \sqrt{(0 - 0)^2 + (0 - 0)^2 + (1 - 0)^2} = \sqrt{1} = 1.$
To find $PQ,$ I use the distance formula between $P$ and $Q:$ $PQ = \sqrt{(1 - 0)^2 + (0 - 0)^2 + (0 - 1)^2} = \sqrt{2}.$
Therefore, the length of the path I want is $d = OP + PQ = 1 + \sqrt{2}.$
The problem asks me to enter $d^2,$ so I square $d:$ $d^2 = (1 + \sqrt{2})^2 = 1 + 2\sqrt{2} + 2 = 3 + 2\sqrt{2}.$