The distance of the point $\mathrm{P}(3,4,4)$ from the point of intersection of the line joining the points $\mathrm{Q}(3,-4,-5), \mathrm{R}(2,-3,1)$ and the plane $2 x+y+z=7$ is

Find the distance from point $\mathrm{P}(3,4,4)$ to the intersection of line $\mathrm{QR}$ with plane $2x+y+z=7$.

A point on line $\mathrm{QR}$ through $\mathrm{Q}(3,-4,-5)$ and $\mathrm{R}(2,-3,1)$ is parametrized as $(3-t, -4+t, -5+6t)$.

Substitute into the plane equation:
$2(3-t) + (-4+t) + (-5+6t) = 7$
$6 - 2t - 4 + t - 5 + 6t = 7$
$-3 + 5t = 7$
$5t = 10$
$t = 2$

The intersection point is $\mathrm{S}(1,-2,7)$ when $t=2$.

The distance from $\mathrm{P}(3,4,4)$ to $\mathrm{S}(1,-2,7)$ is
$\sqrt{(1-3)^2 + (-2-4)^2 + (7-4)^2} = \sqrt{4 + 36 + 9} = \sqrt{49} = 7$

Final answer: $\boxed{7}$ units