It is known that the equation of the line passing through the points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ is$\;\frac{x - x_1}{x_2 - x_1} = \;\frac{y - y_1}{y_2 - y_1} = \;\frac{z - z_1}{z_2 - z_1}$So here we have,$x_1 = 3, y_1 = -4, z_1 = -5 \;\text{ and }\; x_2 = 2, y_2 = -3, z_2 = 1$The line passing through the points, $(3,-4,-5)$ and $(2,-3,1)$ is given by,$\;\frac{x-3}{2-3} = \;\frac{y+4}{-3+4} = \;\frac{z+5}{1+5}$Let $\;\frac{x-3}{-1} = \;\frac{y+4}{1} = \;\frac{z+5}{6} = k$Now,$\Rightarrow \;\frac{x-3}{-1} = k$$\therefore x = -k + 3$$\Rightarrow \;\frac{y+4}{1} = k$$\therefore y = k - 4$ $\Rightarrow \;\frac{z+5}{6} = k$$\therefore z = 6k - 5$The co-ordinates of the point of intersection of the given line and plane is, $(-k+3, k-4, 6k-5)$If is lies on the plane $2x + y + z = 7$, then,$\Rightarrow 2(-k+3) + k - 4 + 6k - 5 = 7$$\Rightarrow 5k = 10$$\therefore k = 2$Substituting k = 2 in ( 1 ) we get,The co-ordinates of the point of intersection of the given line and plane are$(-k+3, k-4, 6k-5)$$\Rightarrow (-2+3, 2-4, 6(2)-5)$$\Rightarrow (1,-2,7)$$\Rightarrow \;\text{Required distance}\; = \;\text{Distance between points}\; (3,4,4) \;\text{ and }\; (1-2,7)$$= \sqrt{(3-1)^2 + (4+2)^2 + (4-7)^2}$$= \sqrt{4+36+9}$ $= \sqrt{49}$ $= 7 \;\text{ units}\;$