The distance of the point 1 , 2 from the line x + y + 5 = 0 measured along the line parallel to 3 x - y = 7…

Let the equation of the line parallel to $3x-y=7$ be $3x-y=\lambda$.

Since, it passes through $\left(1,2\right)$, this point will satisfy the equation $3x-y=\lambda$.

$\therefore 3-2=\lambda$

$\Rightarrow \lambda =1$

So, the line is $3x-y=1$.

Now, $\left(x+y+5\right)+\left(3x-y\right)=0+1\Rightarrow 4x=-4\Rightarrow x=-1$.

Putting in $x+y+5=0,$ we get

$y=-x-5=1-5=-4$

So, the point of intersection of $x+y+5=0$ and $3x-y=1$ is $\left(-1,-4\right)$.

The required distance is the distance between $\left(1,2\right)$ and $\left(-1,-4\right)$.

By distance formula,

the required distance $=\sqrt{{\left\{1-\left(-1\right)\right\}}^2+{\left\{2-\left(-4\right)\right\}}^2}$$=\sqrt{{\left(2\right)}^2+{\left(6\right)}^2}=\sqrt{4+36}=\sqrt{40}$.