Equation of given circle
\(\begin{array}{rlrl}
& x^2+y^2-6 x-10 y+p & =0 \\
\Rightarrow & (x-3)^2+(y-5)^2 & =34-p \\
\therefore & 34-p > 0 \Rightarrow p < 34 \quad \ldots (i)
\end{array}\)
\(\because\) Circle doesn't touch or intersect the coordinate axis, so
\(\begin{array}{ll}
\sqrt{34-p} < 3, \quad \text{for X-axis} \\
\Rightarrow 34-p < 9 \Rightarrow p > 25 \quad \ldots (ii) \\
\text {and } \sqrt{34-P} < 5, \quad \text{for Y-axis} \\
\Rightarrow 34-p < 25 \\
\Rightarrow p > 9 \quad \ldots (iii)
\end{array}\)
\(\because\) Point \((1,4)\) lies inside the circle, so
\(\begin{array}{llll}
& 1+16-6-40+p & < 0 \\
\Rightarrow & p < 29 \quad \ldots (iv)
\end{array}\)
From inequalities Eqs. (i), (ii), (iii) and (iv), we get
\(25 < p < 29\)
Hence, option (b) is correct.