If $X(1,1,1)$ and $Y(2,2,4)$ are two fixed points for $P$ be a variable point in line $2x+y+z=1$.
Then minimum value of $PX+PY.$
Attempt : points $X(1,1,1)$ and $Y(2,2,4)$ are lie on sime side of plane $2x+y+z = 1$
wan,t be able to go further , could some help me
Do you know Heron's problem? If you don't, take a look, and try to apply the ideas here. Below is a solution using this argument
Let $Z$ be the reflection of $Y$ over the plane $\pi: 2x+y+z=1$. First, we'll compute $Z$'s coordinates (i'm not sure if there's a formula to get the coordinates, so we're gonna do it by hand). The plane $\pi$ has normal vector $(2, 1, 1)$, so the line passing by $Y$ and perpendicular to $\pi$ is
$$ l: \left\{(2+2t, 2+t, 4+t) \middle| t \in \mathbb{R} \right\}$$
The intersecton point $M:l \cap \pi$ is easy to get; just plug $l$ coordinates on $\pi$ equation, and solve for $t$. We have
$$ 2(2+2t)+(2+t)+(4+t)=1 \Rightarrow 6t+10=1 \Rightarrow t=-\frac{3}{2} $$
Plugging on $l$'s formula, we get
$$ M =(-1, \frac{1}{2}, \frac{5}{2}) $$
But the middle point of $\overline{ZY}$ is the intersection of the line with the plane, $M$, so
$$ \frac{Z+Y}{2} = M \Rightarrow Z=(-4, -1, 1) $$
Now, for all points $P \in \pi$, we have $\overline{PY}=\overline{PZ}$, because triangles $PMY, PMZ$ are rectangle ($l$ is perpendicular to $\pi$), and thanks to the reflection we have $\overline{MY}=\overline{MZ}$, so Pytagorean Theorem gives us $\overline{PY}^2=\overline{PM}^2+\overline{YM}^2=\overline{PM}^2+\overline{ZM}^2=\overline{PZ}^2$. Thanks to that, minimizing $\overline{PX}+\overline{PY}$ is the same as minimizing $\overline{PX}+\overline{PZ}$.
Let $m$ the line determined by $X$ and $Z$. Of course,
$$ m=\left\{ (1+5t, 1+2t, 1) \middle| t \in \mathbb{R} \right\} $$
and we can get $Q: m \cap \pi$ in the same way as before, getting
$$ 2(1+5t)+(1+2t)+1=1 \Rightarrow 12t+3=0 \Rightarrow t=-\frac{1}{4} $$
so,
$$ Q=\left( -\frac{1}{4}, \frac{1}{2}, 1\right) $$
And we can calculate $\overline{QX}+\overline{QZ}= \sqrt{29}$.For any point $P \in \pi$, using triangle inequallity over $PXZ$ we get
$$ \overline{PX}+\overline{PZ} \geq \overline{XZ}=\overline{XQ}+\overline{QZ}=\sqrt{29} $$
And equallity is achieved when $P=Q$. So, the minimun of $\overline{PX}+\overline{PY}=\overline{PX}+\overline{PZ}$ is $\sqrt{29}$