Find the equation of the smallest sphere which is tangent to both of the lines
$$\begin{pmatrix}
x\\y\\z \end{pmatrix} =\begin{pmatrix}
t+1\\
2t+4\\
-3t +5
\end{pmatrix},\;\;\;\begin{pmatrix}
x\\y\\z \end{pmatrix} =\begin{pmatrix}
4t-12\\
-t+8\\
t+17
\end{pmatrix}.$$
1. **Identify the parametric equations of the lines:**
   \[
   l_1: \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} t+1 \\ 2t+4 \\ -3t+5 \end{pmatrix}
   \]
   \[
   l_2: \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 4t-12 \\ -t+8 \\ t+17 \end{pmatrix}
   \]

2. **Express the lines in vector form:**
   \[
   l_1: \mathbf{r} = \mathbf{a}s + \mathbf{p}, \quad \text{where} \quad \mathbf{a} = \begin{pmatrix} 1 \\ 2 \\ -3 \end{pmatrix}, \quad \mathbf{p} = \begin{pmatrix} 1 \\ 4 \\ 5 \end{pmatrix}
   \]
   \[
   l_2: \mathbf{r} = \mathbf{b}t + \mathbf{q}, \quad \text{where} \quad \mathbf{b} = \begin{pmatrix} 4 \\ -1 \\ 1 \end{pmatrix}, \quad \mathbf{q} = \begin{pmatrix} -12 \\ 8 \\ 17 \end{pmatrix}
   \]

3. **Find the line of shortest distance between \( l_1 \) and \( l_2 \):**
   The line \( l \) of shortest distance between \( l_1 \) and \( l_2 \) is perpendicular to both \( l_1 \) and \( l_2 \). Let:
   \[
   \mathbf{c} = (as + \mathbf{p}) - (bt + \mathbf{q}) = \begin{pmatrix} s - 4t + 13 \\ 2s + t - 4 \\ -3s - t - 12 \end{pmatrix}
   \]

4. **Set up the perpendicularity conditions:**
   \[
   \mathbf{c} \cdot \mathbf{a} = 0 \quad \text{and} \quad \mathbf{c} \cdot \mathbf{b} = 0
   \]
   \[
   \begin{pmatrix} s - 4t + 13 \\ 2s + t - 4 \\ -3s - t - 12 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ 2 \\ -3 \end{pmatrix} = 0
   \]
   \[
   (s - 4t + 13) + 2(2s + t - 4) - 3(-3s - t - 12) = 0
   \]
   Simplifying:
   \[
   s - 4t + 13 + 4s + 2t - 8 + 9s + 3t + 36 = 0
   \]
   \[
   14s + t + 41 = 0 \quad \Rightarrow \quad 14s + t = -41 \quad \text{(Equation 1)}
   \]

   Similarly,
   \[
   \begin{pmatrix} s - 4t + 13 \\ 2s + t - 4 \\ -3s - t - 12 \end{pmatrix} \cdot \begin{pmatrix} 4 \\ -1 \\ 1 \end{pmatrix} = 0
   \]
   \[
   4(s - 4t + 13) - (2s + t - 4) + (-3s - t - 12) = 0
   \]
   Simplifying:
   \[
   4s - 16t + 52 - 2s - t + 4 - 3s - t - 12 = 0
   \]
   \[
   -s - 18t + 44 = 0 \quad \Rightarrow \quad s + 18t = 44 \quad \text{(Equation 2)}
   \]

5. **Solve the system of linear equations:**
   \[
   \begin{cases}
   14s + t = -41 \\
   s + 18t = 44
   \end{cases}
   \]
   Multiply the second equation by 14:
   \[
   14s + 252t = 616
   \]
   Subtract the first equation from this result:
   \[
   14s + 252t - (14s + t) = 616 - (-41)
   \]
   \[
   251t = 657 \quad \Rightarrow \quad t = \frac{657}{251}
   \]
   Substitute \( t \) back into the first equation:
   \[
   14s + \frac{657}{251} = -41
   \]
   \[
   14s = -41 - \frac{657}{251}
   \]
   \[
   14s = -\frac{10291}{251} \quad \Rightarrow \quad s = -\frac{10291}{3514} = -\frac{782}{251}
   \]

6. **Find the center of the sphere:**
   Substitute \( s \) and \( t \) back into the parametric equations of \( l_1 \) and \( l_2 \):
   \[
   \mathbf{r}_1 = \begin{pmatrix} -\frac{782}{251} + 1 \\ 2 \left( -\frac{782}{251} \right) + 4 \\ -3 \left( -\frac{782}{251} \right) + 5 \end{pmatrix} = \begin{pmatrix} -\frac{531}{251} \\ -\frac{1313}{251} \\ \frac{2346}{251} \end{pmatrix}
   \]
   \[
   \mathbf{r}_2 = \begin{pmatrix} 4 \left( \frac{657}{251} \right) - 12 \\ -\frac{657}{251} + 8 \\ \frac{657}{251} + 17 \end{pmatrix} = \begin{pmatrix} \frac{1311}{251} \\ \frac{1343}{251} \\ \frac{4914}{251} \end{pmatrix}
   \]
   The center of the sphere is the midpoint of \( \mathbf{r}_1 \) and \( \mathbf{r}_2 \):
   \[
   \mathbf{C} = \frac{1}{2} \left( \begin{pmatrix} -\frac{531}{251} \\ -\frac{1313}{251} \\ \frac{2346}{251} \end{pmatrix} + \begin{pmatrix} \frac{1311}{251} \\ \frac{1343}{251} \\ \frac{4914}{251} \end{pmatrix} \right) = \begin{pmatrix} -\frac{915}{502} \\ \frac{791}{502} \\ \frac{8525}{502} \end{pmatrix}
   \]

7. **Find the radius of the sphere:**
   The radius is half the distance between \( \mathbf{r}_1 \) and \( \mathbf{r}_2 \):
   \[
   \text{Distance} = \sqrt{ \left( \frac{1842}{251} \right)^2 + \left( \frac{2656}{251} \right)^2 + \left( \frac{2568}{251} \right)^2 }
   \]
   \[
   = \frac{1}{251} \sqrt{1842^2 + 2656^2 + 2568^2} = \frac{1}{251} \sqrt{3395364 + 7054336 + 6594624} = \frac{1}{251} \sqrt{17044324}
   \]
   \[
   = \frac{1}{251} \times 4130 = \frac{4130}{251}
   \]
   The radius is:
   \[
   \frac{4130}{502} = \frac{2065}{251}
   \]

The final answer is \(\boxed{\left( -\frac{915}{502}, \frac{791}{502}, \frac{8525}{502} \right), \frac{2065}{251}}\)