Minimum value of $PA+PB$ is If  $P(x,y,z)$ lie on line $\displaystyle \frac{x+2}{2}=\frac{y+7}{2}=\frac{z-2}{1}$ and $A(5,3,4)$ and $B(1,-1,2)$ . Then minimum value of $PA+PB$ is 
what i try
let $\displaystyle \frac{x+2}{2}=\frac{y+7}{2}=\frac{z-2}{1}=\lambda$
Then $P(2\lambda-2,2\lambda-7,\lambda+2)$ 
$PA+PB=\sqrt{(2\lambda-7)^2+(2\lambda-10)^2+(\lambda-2)^2}+\sqrt{(2\lambda-3)^2+(2\lambda-8)^2+\lambda^2)}$
$PA+PB=\sqrt{9\lambda^2-72\lambda+153}+\sqrt{
9\lambda^2-36\lambda+45}$
how do i minimize it b3cause derivative method is very tedious  help me please

By Minkowski we obtain:
$$\sqrt{9\lambda^2-72\lambda+153}+\sqrt{
9\lambda^2-36\lambda+45}=$$
$$=3\left(\sqrt{(4-\lambda)^2+1}+\sqrt{(\lambda-2)^2+1}\right)\geq$$
$$\geq3\sqrt{(4-2)^2+(1+1)^2}=6\sqrt2.$$
The equality occurs for $\lambda=3,$ which says that we got a minimal value.
