Equation of the plane containing the line $\frac{x-1}{1}=\frac{y-2}{2}=\frac{z-3}{3}$ is
$$
a(x-1)+b(y-2)+c(z-3)=0
$$


where $a .1+b .2+c .3=0$
i.e., $\quad a+2 b+3 c=0$
(ii)
Since the plane (i) parallel to the line
$$
\begin{array}{ll}
& \frac{x}{1}=\frac{y}{1}=\frac{z}{4} \\
\therefore \quad & a .1+b .1+c .4=0 \\
\text { i.e., } \quad & \mathrm{a}+\mathrm{b}+4 \mathrm{c}=0
\end{array}
$$
(iii)
From (ii) and (iii),
$$
\begin{gathered}
\frac{a}{8-3}=\frac{b}{3-4}=\frac{c}{1-2}=k \quad \text { (let) } \\
\therefore \quad a=5 k, b=-k, c=-k
\end{gathered}
$$
On putting the value of $a, b$ and $c$ in equation (i),
$$
\begin{aligned}
&5(x-1)-(y-2)-(z-3)=0 \\
&\Rightarrow 5 x-y-z=0
\end{aligned}
$$
when $x=1, y=0$ and $z=5$; then
L.H.S. of equation (iv) $=5 x-y-2$
$$
\begin{aligned}
&=5 \times 1-0-5 \\
&=0 \\
&=\text { R.H.S. of }
\end{aligned}
$$
equation (iv)
Hence coordinates of the point $(1,0,5)$ satisfy the equation plane represented by equations (iv),
Therefore the plane passes through the point $(1,0,5)$