How to find a closest point on a constraint to a given point? (This is a multi-variable calculus problem, not a linear algebra one)
Find the closest point to the point $(2,7,8)$ on the constraint:
$$ 4x + 7y = x+3y+5z$$
And find the projection of the vector $(2,7,8)$ on the plane:
$$3x+4y =5z$$
My try:
I thought to use Lagrange multipliers, but then switched to just finding the line perpendicular to the plane (Which I am not sure if is a legal move) :
$g(x,y,z) = 3x + 4y - 5z = 0$
The vector perpendicular to the plane is $\vec{n} = (3,4,-5)$
and so the line that is perpendicular to the plane, and goes through the point is: $(2,7,8) + t(3,4,-5) = (2+3t, 7+4t, 8-5t)$
And if we look where it intersects the plane (then it has to satisfy the plane's equation):
$ 3(2+3t) + 4(7+4t) -5(8-5t) = 0 \Rightarrow 50t = 6 \Rightarrow t = \frac{3}{25}$
And thus the closest point is: $(2 + \frac{9}{25}, 7+\frac{12}{25}, 8-\frac{30}{25}) = ( 2.36, 7.48, 7.4)$
For the projection, I need to find $P_w( \vec{(2,7,8)} ) $ and the projection will be: $\vec{(2,7,8)} - P_w( \vec{2,7,8})$ But couldn't find $P_w$ I thought it is just $xy$ component $(2,7,0)$ and then we negate the vector $(-2,-7,-8)$ and add them to get: $(0,0, -8)$
 A: Your calculation of the closest point is correct.
Concerning the second question: If $u$ is a vector and $v$ is a second vector (both non-zero), then the projection $\operatorname{proj}_v u$ of $u$ onto $v$ is given by
$$\operatorname{proj}_v u = \frac{\langle u, v\rangle}{\|v\|^2} v.$$
Now, a moment's thought reveals that the projection of $u$ onto a plane $E$ with normal vector $n$ is given by
$$\operatorname{proj}_E u = u - \operatorname{proj}_n u.$$
(To imagine this, think of the plane to be horizontal. Then calculating the projection means computing $u$ minus the vertical component of $u$. What remains then is the horizontal component. The vertical component is exactly given by the projection of $u$ onto $n$.)
Plugging in the given values with $u = (2, 7, 8)^\intercal$ and $n = (3, 4, -5)^\intercal$ finally shows that the results are indeed the same:
$$\begin{pmatrix} 2 \\ 7 \\ 8 \end{pmatrix} - \frac{-6}{50} \begin{pmatrix} 3 \\ 4 \\ -5 \end{pmatrix} = \begin{pmatrix} 2 + 18/50 \\ 7 + 24/50 \\ 8 - 30/50 \end{pmatrix}.$$
