Answer:
Option A
Explanation:
We have,
OA=$\hat{i}+2\hat{j}+3\hat{k}$ and OB= $4 \hat{i}+\hat{k}$
$\because$ $OA \times OB=\begin{bmatrix}\hat{i} & \hat{j} & \hat{k} \\1 & 2&3 \\4 &0&1 \end{bmatrix}= 2 \hat{i}+11\hat{j}-8\hat{k}$
equation of line passing through B and parallel to OA x OB is
r= $4\hat{i}+\hat{k}+\lambda (2 \hat{i}+11 \hat{j}-8\hat{k})$
r= $(4+2 \lambda)\hat{i}+11 \lambda \hat{j}+(1-8 \lambda) \hat{k}$
Distance from B to line is $\sqrt {189}$
$\therefore$ $\sqrt{189}$= $\sqrt{(4+2\lambda-4)^{2}+(11\lambda)^{2}+(1-8 \lambda -1)^{2}}$
189= $4\lambda^{2}+121\lambda^{2}+64 \lambda^{2}$
189= 189 $\lambda ^{2}$
$\lambda$= $\pm$ 1
$\therefore$ Position vector ;ie on line B
r= $6 \hat{i}+11\hat{j}-7 \hat{k}$
