The shortest distance between given lines $\mathbf{r}=(3 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}-2 \hat{\mathbf{k}})+t(-\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+\hat{\mathbf{k}})$ and $\mathbf{r}=(\hat{\mathbf{i}}-7 \hat{\mathbf{j}}-2 \hat{\mathbf{k}})+s(\hat{\mathbf{i}}+3 \hat{\mathbf{j}}+2 \hat{\mathbf{k}})$ is
$\begin{aligned} d & =\frac{\left|\begin{array}{ccc}2 & 11 & 0 \\ -1 & 2 & 1 \\ 1 & 3 & 2\end{array}\right|}{\sqrt{(4-3)^2+(-2-1)^2+(-3-2)^2}} \\ & =\frac{|2(4-3)-11(-2-1)|}{\sqrt{1+9+25}}=\frac{35}{\sqrt{35}}=\sqrt{35}\end{aligned}$
Now the projection of $\mathbf{p}=-2 \hat{\mathbf{i}}+11 \hat{\mathbf{j}}$ on vector $\mathbf{Q}=-\hat{\mathbf{i}}+3 \hat{\mathbf{j}}+5 \hat{\mathbf{k}}$ is also $\sqrt{35}$
In option (c) inside of $\hat{\mathbf{i}}+3 \hat{\mathbf{j}}+5 \hat{\mathbf{k}}$ put $-\hat{\mathbf{i}}+3 \hat{\mathbf{j}}+5 \hat{\mathbf{k}}$ then option (c) is ok