Orthogonal projection of a line onto a plane Let $\pi:x+y+z=1$ describe a plane and let $L$ be a line s.t. $L:(x,y,z)= (1,1,-1)+t(1,0,1)$, $t\in\mathbb{R}$. When $L$ is orthogonally projected onto $\pi$ a new line is made, write this new line in scalar form.
I know how to write a line of an orthogonal projection of a point onto a plane, but here I'm not sure what to do and I don't think I understand the question. That $L$ is orthogonally projected onto $\pi$, I understand as every point on $L$ is orthogonally projected onto $\pi$ so there will be infinite many new lines not just one.

"I understand as every point on $L$ is orthogonally projected onto $\pi$ so there will be infinite many new lines not just one"
We claim that the orthogonally projected points are all along the same line.
The projection of $P_t=(1,1,−1)+t(1,0,1)$, which belongs to the line $L$, onto the plane $x+y+z=1$ is given by the intersection of the plane and the line $s\to P_t+(1,1,1)s$, which is orthogonal to $\pi$:
$$(1+t+s)+(1+s)+(-1+t+s)=1\implies s=-\frac{2t}{3}.$$
Hence the orthogonal projection of $P_t$ onto $\pi$ is
$$Q_t=P_t-\frac{2t}{3}(1,1,1)=(1,1,−1)+t(1,0,1)-\frac{2t}{3}(1,1,1)=
(1,1,−1)+\frac{t}{3}(1,-2,1).
$$
Notice that any projected point $Q_t$ lays along the same line: $t\to Q_t=(1,1,−1)+\frac{t}{3}(1,-2,1).$
