If a is a vector of magnitude 50 , collinear with the vector b =6 i...

Since $\vec{a}=m \vec{b}$ for some scalar $m$ i.e., $$\begin{array}{l} \vec{a}=m\left(6 \hat{i}-8 \hat{j}-\frac{15}{2} \hat{k}\right) \\ \Rightarrow|a|=|m| \sqrt{36+64+\frac{225}{4}} \\ \Rightarrow 50=\frac{25}{2}|m| \\ \Rightarrow|m|=4 \\ \Rightarrow m=\pm 4 \end{array}$$Since, a makes an acute angle with the positive direction of $Z$ - axis, so its $z$ component must be positive and hence, $m$ must be $-4$. \therefore a=-4\left(6 \hat{i}+8 \hat{j}-\frac{15}{2} \hat{k}\right)=-24 \hat{i}+32 \hat{j}+30 \hat{k}$