Equation of line joining points $(\hat{\mathbf{i}}+\hat{\mathbf{j}})$ and $(3 \hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}})$ in cartesian form is
$$
\frac{x-1}{2}=\frac{y-1}{0}=\frac{z}{-1}=r(\text { let })
$$
Then, point on line Eq. (i) is $p(2 r+1,1,-r)$.
Now, equation of plane passes through the point $(2 \hat{\mathbf{i}}+4 \hat{\mathbf{j}})$ and parallel to the vectors $(3 \hat{\mathbf{j}}+5 \hat{\mathbf{k}})$ and $(3 \hat{\mathbf{i}}-\hat{\mathbf{k}})$ in cartesian form is
$$
\begin{aligned}
& \left|\begin{array}{ccc}
x-2 & y-4 & z \\
0 & 3 & 5 \\
3 & 0 & -1
\end{array}\right| & =0 \\
\Rightarrow & -3(x-2)+15(y-4)-9 z & =0 \\
\Rightarrow & 3 x-15 y+9 z+54 & =0
\end{aligned}
$$
Let the point $P(2 r+1,1,-r)$ on the plane Eq. (ii) itself, so $r=14$
So, position vector of the point $P$ is
$$
\mathbf{O P}=29 \hat{\mathbf{i}}+\hat{\mathbf{j}}-14 \hat{\mathbf{k}} .
$$