Answer:
Option D
Explanation:
Given P(4,5,x), Q (3,y,4) and R(5,8,0) are collinear
$\therefore$ $\overrightarrow{PQ}=\lambda \overrightarrow{QR}$
$\Rightarrow$ $ \overrightarrow{PQ}=\overrightarrow{OQ}-\overrightarrow{OP}$
= $(3\hat{i}+y\hat{j}+4\hat{k})-(4\hat{i}+5\hat{j}+x\hat{k})=-\hat{i}+(y-5)\hat{j}+(4-x)\hat{k}$
and
$\overrightarrow{QR}=\overrightarrow{OR}-\overrightarrow{OQ}=(5\hat{i}+8\hat{j})-(3\hat{i}+y\hat{j}+4\hat{k})= 2\hat{i}+(8-y)\hat{j}-4\hat{k}$
$\therefore$ $-\hat{i}+(y-5)\hat{j}+(4-x)\hat{k}$
= $\lambda[2\hat{i}+(8-y)\hat{j}-4\hat{k}]$
On equating the component of vector both sides, we get,
$\frac{-1}{2}=\lambda,\frac{y-5}{8-y}=\lambda,\frac{4-x}{-4}=\lambda$
On putting the value of $\lambda$ , we get
$(y-5)=(8-y)(-\frac{1}{2})$
and $(4-x)=-4(-\frac{1}{2})$
$\Rightarrow$ 2y-10=y-8
and 4-x=2
$\Rightarrow$ y=2
and x=2
$\therefore$ x+y=2+2=4
