Coordinate Axes?

Let equation of line be $\frac xa + \frac yb =1$

Then length of intercept cut on axis will be $\sqrt{a^2 + b^2} = \sqrt5$

$\rightarrow a^2 + b^2 = 5 ....... (i)$

Also $(2,2)$ lies on the line so we get $(a+b) = \frac{ab}{2}$

Squaring both sides we get $a^2 +b^2 +2ab = \frac{a^2\times b^2}{4}$

Substituting value of $a^2 + b^2$

This forms a quadratic equation in $ab$ whose roots are $-2 ,10$ , we get $2ab = -4 , 20$

Adding and subtracting with eq(i) we get

$a+b = \pm1$ and $a-b = \pm3$ ( we will not take $ab = 10$ as this makes $(a-b)^2$ imaginary

We get our equation as:

$\frac{x}{\pm2} + \frac{y}{\pm1} = 1$

Now to get our equation we put $(2,2)$ . This gives the equation as $x - 2y + 2 =0$

So our answer is $1^3 + (-2)^2 + 2 = \boxed{7}$

P.S I am getting different answer as yours because i took $(a-b)^2 = 9$ not $(b-a)^2 = 9$ . This therefore changes the answer but taking the other case you can get your required answer. This question is pretty sign sensitive If we take equation to be $-x + 2y -2 =$ we get answer $1$ instead of $7$

The answer is:

$2x - y - 2 = 0$

Please do post your solution if you have solved it. It will be very much appreciated. Thank you. :)