The distance of the point P 3 , 4 , 4 from the point of intersection of the line joining the points Q 3 , -…

We have, point $P\left(3,4,4\right)$

Equation of line joining the points $Q\left(3,-4,-5\right)$ and $R\left(2,-3,1\right)$ is $\frac{x-3}{-1}=\frac{y+4}{1}=\frac{z+5}{6}=r$

Any point on above line is $\left(-r+3,r-4,6r-5\right)$

Now, satisfying it in the given plane $2x+y+z=7,$ we get

$2\left(-r+3\right)+\left(r-4\right)+\left(6r-5\right)=7$

$\Rightarrow r=2$

So, required point of intersection is $T\left(1,-2,7\right).$

Hence, $PT=\sqrt{{\left(3-1\right)}^2+{\left(4+2\right)}^2+{\left(4-7\right)}^2}=7.$