Unit vector along the direction of vector $\overrightarrow{\mathit{a}}$ , $\overrightarrow{\mathit{a}}$ = $\frac{\overrightarrow{\mathit{a}}}{\left|\overrightarrow{\mathit{a}}\right|}$
Let $\overrightarrow{\mathit{a}}$ = $\hat{\mathit{i}}-2\hat{\mathit{j}}+2\hat{\mathit{k}}$
$\left|\overrightarrow{\mathit{a}}\right|$ = $\sqrt{{\left(1\right)}^2+{\left(-2\right)}^2+{\left(2\right)}^2}$ = ± 3
i.e. $\hat{\mathit{a}}$ = $\frac{1}{3}\left(\hat{\mathit{i}}-2\hat{\mathit{j}}+2\hat{\mathit{k}}\right)$
So, the vector whose magnitude is 15 and has direction along the vector $\hat{\mathit{i}}-2\hat{\mathit{j}}+2\hat{\mathit{k}}$ is given by,
15 × $\left(\frac{1}{3}\right)\left(\hat{\mathit{i}}-2\hat{\mathit{j}}+2\hat{\mathit{k}}\right)$
= 5 $\left(\hat{\mathit{i}}-2\hat{\mathit{j}}+2\hat{\mathit{k}}\right)$
= $\left(5\hat{\mathit{i}}-10\hat{\mathit{j}}+10\hat{\mathit{k}}\right)$
So the required vector is $\left(5\hat{\mathit{i}}-10\hat{\mathit{j}}+10\hat{\mathit{k}}\right)$