Find the projection of the point on the line

Question

Solve the equation of the projection of the point $A(1,2,8)$ on the straight line $p$ with the property:

$$p=\frac{x-1}{2}=\frac{y}{-1}=\frac{z}{1}.$$

Answer 1

HINTS

1. Find the coordinates of $p$; $\frac{x-1}{2}=\frac{y}{-1}$ means $1-x=2y$, but $1-x=-2z$, so $y=-z$, so $p=(1-2a,a,-a)$, which means the equations given do not uniquely deterimine a line!
2. In any case, provided you have coordinates, $q$ is the projection of $A$ onto the line of $p$ iff one has both $q=tp$ with $t\in\mathbb{R}$ and $A-q$ is orthogonal to $p$.

Leaving the details to you.

Edit

I now see $p$ is a line, not a point generating the line. In this case, the line is $(1,0,0)-a(2,-1,1)$. So for point 2., we have $q=(1,0,0)+a(2,-1,1)$ and $A-q=(1,2,8)-(1,0,0)-a(2,-1,1)=(0,2,8)-a(2,-1,1)$ perpendicular to $(2,-1,1)$. Which means:

$$(0,2,8)\cdot(2,-1,1)-a(2,-1,1)\cdot(2,-1,1)=0,$$

or:

$$-2+8-a(4+1+1)=0,$$

that is:

$$a=\frac{8-2}{4+1+1}=\frac{6}{6}=1.$$

So the desired point will be $q=(1,0,0)+(2,-1,1)=(3,-1,1)$.

Answer 2

Notice,

Let the projection of the point $A(1, 2, 8)$ be $P(2p+1, -p, p)$ on the given line: $p=\frac{x-1}{2}=\frac{y}{-1}=\frac{z}{1}$

Then $\vec{AP}=(2p, -p-2, p-8)$ will be normal to the vector $(2, -1, 1)$ (parallel to the line) then the dot product should be zero $$2(2p)+(-1)(-p-2)+(1)(p-8)=0\implies p=1$$ setting $p=1$, the point of projection is $\color{red}{P(3, -1, 1)}$