Let the line joining the points with position vectors $-2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+5 \hat{\mathbf{k}}$ and $7 \hat{\mathbf{i}}-\hat{\mathbf{k}}$ be divide in the ratio $\lambda: 1$ by $\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}$.
$\begin{aligned} & \therefore \frac{\lambda(7 \hat{\mathbf{i}}-\hat{\mathbf{k}})+(-2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+5 \hat{\mathbf{k}})}{\lambda+1}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}} \\ & \Rightarrow \quad(7 \lambda-2) \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+(5-\lambda) \hat{\mathbf{k}}=(\lambda+1) \hat{\mathbf{i}} \\ & +2(\lambda+1) \hat{\mathbf{j}}+3(\lambda+1) \hat{\mathbf{k}} \\ & \end{aligned}$
On equating the coefficient of $\mathbf{i}$, we get
$\begin{aligned} & \quad \begin{aligned} & 7 \lambda-2=\lambda+1 \\ & \Rightarrow \quad \lambda \lambda=2 \\ & \text { Hence, ratio } \lambda: 1=2: 1 .\end{aligned}\end{aligned}$