What we have is a line with positive slope, passing through the point $(-2, 5)$, with the length of the segment on this line from the point where it intersects the negative $x$-axis at $(x_0, 0)$ and the positive $y$-axis at $(0, y_0)$ equal to $7\sqrt 2$.
We know then, that $$x_0^2 + y_0^2 = (7\sqrt 2)^2 = 98\tag{1}$$
We also know that the distance from $(x_0, 0)$ to $(-2, 5)$ plus the distance from $(-2, 5)$ to the point $(0, y_0)$ is equal to $7\sqrt 2$: $$\sqrt {25 + (x_0+2)^2} + \sqrt{4+(y_0-5)^2} = 7\sqrt 2\tag{2}.$$
Solving $(1)$ and $(2)$ simultaneously gives us the real solutions $x_0 = -7$, and $y_0 = 7$. With these values, we must have that the slope of each of these "segments" comprising the entire length are equal: $$m = \frac{5-0}{-2 - x_0} = \frac{5-y_0}{-2-0}= 1$$
Now, given your point on the line $(-2, 5)$ and the slope of $m = 1$, we can construct the equation of the line: $$y - 5 = x + 2 \iff y = x+7$$
For a parametric equation of a line $(x,y,z) = (x_0, y_0, z_0) + (a,b,c)s$, the line itself is the set of all points $P(x,y,z)$ such that $x = x_0 + as$, $y = y_0 + bs$, and $z = z_0 + cs$. When $s = 1$, this gives you one point $A$; when $s = 2$, this gives you another point $B$; and when you take all values of $s \in \mathbb{R}$, you get the entire line.
Notice that if instead you had $(x,y,z) = (x_0, y_0, z_0) + (a,b,c)s/2$, you'd still have the same line, except this time, $s$ has to be $2$ to give you point $A$, $s = 4$ gives you point $B$, etc. So when you're trying to find the value of $s$ for any one point, you can just choose any $s$ that you want!
So if $(2,1,1) = (a,b,c)s$, choose any $s$ you want and solve it for $a$, $b$, and $c$.
(By the way, there are two points on the line $(1,2,0) + (2,-1,2)t$ that are $3$ units away from $(1,2,0)$. You found one when you set $t = 1$; what if you set $t = -1$?)
Best Answer
hint.....The point of intersection is $(13,8)$. A line of gradient $m$ through this point is $$y-8=m(x-13)\Rightarrow y-mx+13m-8=0$$
Now you can use the formula for the distance $d$ from a point $(p,q)$ to the line $ax+by+c=0$ which is $$d=\left|\frac{ap+bq+c}{\sqrt{a^2+b^2}}\right|$$ You can now form and solve a quadratic equation to find the possible values of $m$