all you need to do is measure how the vectors vec3(1,0,0) vec3 (0,1,0) vec3(0,0,1) get changed by any matrix.
vec3(1,0,0)'s transformation is equal to the first column of the matrix
vec3(0,1,0)'s transformation is equal to the second column of the matrix
vec3(0,0,1)'s transformation is equal to the second column of the matrix (only if it has 3 dimensions)
vec4(0,0,0,1)'s transformation is equal to the second column of the matrix (only if it has 4 dimensions),...
yes, learn how matrices function on basic core level:
in a 2d to 2d matrix transformation, 2 matrix colums are equal to what the 2 base-vectors vec2(1,0) and vec2(0,1) are changed into by the matrix. any scalar of that is also transformed linearily scaled to that, as a matrix transformation is a linear transformation.
in a 3d to 3d matrix transformation, same goes for 3 base-vectors vec3(1,0,0) vec3 (0,1,0) vec3(0,0,1) being changed by 3 matrix columns.
in a 3d to 2d matrix, 3 columns of 2 lines tell how 3 3d basis vectors as in the above are scamed in 3d space, as a sum of three 2d vectors (that get scaled by the matrix).
Best Answer
"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: $t\to Q_t=(1,1,−1)+\frac{t}{3}(1,-2,1).$