Here A(0, 0, 6), B(0, 4, 0) and C(6, 0, 0) are the vertices of ΔABC. Now D is mid-point of BC. ∴ Coordinates of D are $\left(\frac{0+6}{2},\frac{4+0}{2},\frac{0+0}{2}\right)$ = (3 , 2 , 0) ∴ AD = $\sqrt{(0-3)^2+(0-2)^2+(6-0)^2}$ = $\sqrt{9+4+36}$ = 7 units Also E is mid-point of AC ∴ Coordinates of E are $\left(\frac{0+6}{2},\frac{0+0}{2},\frac{6+0}{2}\right)$ = (3 , 0 , 3) ∴ BE = $\sqrt{(0-3)^2+(4-0)^2+(0-3)^2}$ = $\sqrt{9+16+9}$ = $\sqrt{34}$ units Also F is mid point of AB ∴ Coordinates of F are $\left(\frac{0+0}{2},\frac{0+4}{2},\frac{6+0}{2}\right)$ = (0 , 2 , 3) ∴ CF = $\sqrt{(6-0)^2+(0-2)^2+(0-3)^2}$ = $\sqrt{36+4+9}$ = 7 units