Projection vectors

Question

So I have done part of this question. Using the projection vector equation, I found that the projection vector of (4,2,-1) onto V was: (1,-1,-1)

And a vector orthogonal to V is: (3,3,0)=(4,2,-1)-(1,-1,-1) But, just not sure how to find the point in V which is closest to (4,2,-1). Any help????

Answer 1

HINT

Note that $(3,3,0)$ is orthogonal to the plane

$$(4,2,-1)+t(3,3,0)$$

for some $t$ belongs to it.