(d) Let the point $\mathrm{A}$ is $(\alpha, \beta, \gamma)$ and line $\mathrm{L}$ parallel to the vector
Required line $\mathrm{L}=\vec{r}=(\alpha \hat{i}+\beta \hat{j}+\gamma \hat{k})+\lambda(2 \hat{i}+\hat{j}-2 \hat{k})$
Here, $\overrightarrow{\mathrm{OP}}=\vec{r}=-7 \hat{i}-5 \hat{j}+11 \hat{k}$.
Now, $(-7 \hat{i}-5 \hat{j}+11 \hat{k})=(\alpha \hat{i}+\beta \hat{j}+\gamma \hat{k})+\lambda(2 \hat{i}+\hat{j}-2 \hat{k})$
$(-7 \hat{i}-5 \hat{j}+11 \hat{k})=\hat{i}(\alpha+2 \lambda)+\hat{j}(\beta+\lambda)+\hat{k}(\gamma-2 \lambda)$
$\begin{aligned}
& \alpha+2 \lambda=-7,-5=\beta+\lambda, \gamma-2 \lambda=11 \\
& \alpha=-7-2 \lambda, \beta=(-5-\lambda), \gamma=(11+2 \lambda)
\end{aligned}$
Here, $|\overrightarrow{\mathrm{AP}}|=12$
$\overline{\mathrm{AP}}=\hat{i}(-7-\alpha)+\hat{j}(-5-\beta)+\hat{k}(11-\gamma)$
Take modulus both sides, then
$12=\sqrt{(-7-\alpha)^2+(-5-\beta)^2+(11-\gamma)^2}$
Take square both sides,
$\begin{aligned}
& (12)^2=(-7-(-2 \lambda-7))^2+(-5-(-5-\lambda))^2+(11-11-2 \lambda)^2 \\
& 144=4 \lambda^2+\lambda^2+4 \lambda^2 \\
& 9 \lambda^2=144 \\
& \lambda^2=16 \\
& \lambda= \pm 4
\end{aligned}$
Put the value of $\lambda$ in the values of $\alpha, \beta, ~\&~ \gamma$.
$\begin{aligned}
& \alpha=-7-2 \times 4=-7-8=-15 \\
& \beta=-5-4=-9 \\
& \gamma=11+2 \lambda=11+8=19
\end{aligned}$
So, $\overline{\mathrm{OA}}=\alpha \hat{i}+\beta \hat{j}+\gamma \hat{k}=-15 \hat{i}-9 \hat{j}+19 \hat{k}$.