The position vector of the foot of the perpendicular drawn from the point $2\hat{i} - \hat{j} + 5\hat{k}$ to the line $\vec{r}= (11\hat{i} - 2\hat{j}- 8\hat{k})+ λ(10\hat{i} - 4\hat{j}- 11\hat{k})$, is

$-\hat{i} + 3\hat{j}+2\hat{k}$

Let L be the foot of the perpendicular drawn from $P(2\hat{i}-\hat{j} + 5\hat{k})$ on the line

$\vec{r} = 11\hat{i} - 2\hat{j} - 8 \hat{k} + λ (10\hat{i} - 4\hat{j} - 11\hat{k})$.

Let the position vector of L be

$11\hat{i} - 2\hat{j} - 8 \hat{k} + λ (10\hat{i} - 4\hat{j} - 11\hat{k})$

$=(11+10λ)\hat{i} + (-2-4λ)\hat{j} +(- 8 -11λ)\hat{k} $

Then,

$⇒ \vec{PL}= ( 9 + 10 λ) \hat{i} + (-1-4λ) \hat{j} + (-13 -11λ) \hat{k}$

Since PL is perpendicular to the given line which is parallel to vector $\vec{b} = 10 \hat{i} - 4\hat{j} - 11\hat{k}.$

$∴\vec{PL}.\vec{b}= 0 $

$⇒ 10(9 + 10 λ) -4(-1-4 λ) -11(-13-11 λ)= 0  ⇒  λ= - 1$

Putting the value of  λ, we obtain the position vector of L as $\hat{i} + 2\hat{j} + 3\hat{k}.$