If the point (3,4) lies on the locus of the point of intersection of the lines 

 $x\cos \alpha +y\sin \alpha =a$ and$x\sin \alpha -y\cos \alpha =b\left(\alpha \text{\hspace{0.17em}\hspace{0.33em}is\hspace{0.17em}\hspace{0.17em}a\hspace{0.17em}variable}\right)$ ,

The point (a,b) lies on the line $3x-4y=0$ then $\left|a+b\right|$ is equal to 

Squaring and adding the given equations of the lines we get $x^2+y^2=a^2+b^2$  as the locus of the point of intersection of these lines. 
Since (3,4) lies on the locus, we get 
$9+16=a^2+b^2$  i.e   $a^2+b^2=25 ...\left(i\right)$ 
Also, $(a,b)$   lies on  $3x-4y=0$
so     $3a-4b=0$   $\Rightarrow b=(3/4)a ...\left(ii\right)$
From (i) ,   $a^2+(9/16)a^2=25\Rightarrow$
So that   ${\left|a+b\right|}^2={(7/4)}^2{a}^2=49$ $\left|a+b\right|=7$