Let the position vectors of points P ,   Q ,   R and S be a → = i ^ + 2 j ^ - 5 k ^ , b…

Given,

$P(1,2,-5), Q(3,6,3), R\left(\frac{17}{5},\frac{16}{5},7\right) \& S(2,1,1)$

Now solving options,

Option $\left(A\right)$ checking points are coplanar,

$\overrightarrow{PQ}=\left(2,4,8\right), \overrightarrow{PR}=\left(\frac{12}{5},\frac{6}{5},12\right) \& \overrightarrow{PQ}=\left(1,-1,6\right)$

Now if points are coplanar then, $\left[\begin{array}{ccc}\overrightarrow{PQ}& \overrightarrow{PR}& \overrightarrow{PS}\end{array}\right]=0$

Now solving L.H.S, we get,

$\left|\begin{array}{ccc}2& 4& 8\\ \frac{12}{5}& \frac{6}{5}& 12\\ 1& -1& 6\end{array}\right|=\frac{2}{5}\left|\begin{array}{ccc}1& 2& 4\\ 12& 6& 60\\ 1& -1& 6\end{array}\right|=\frac{2}{5}\left(96-24-72\right)=0$

Hence, they are coplanar

So option $\left(A\right)$ is wrong. 

Now solving, option $\left(B\right)$ we get,

$\frac{\overrightarrow{b}+2\overrightarrow{d}}{3}=\frac{7i+8j+5k}{3}$

Now let point divides the $PR \text{in} \lambda :1$ we get,

So, on taking coefficient of $\hat{i}$ we get,

$\Rightarrow \frac{\frac{17\lambda }{5}+1}{(\lambda +1)}=\frac{7}{3}$

$\Rightarrow \frac{17\lambda }{5}+1=\frac{7}{3}(\lambda +1)$

$\Rightarrow 51\lambda +15=35\lambda +35$

$\Rightarrow 16\lambda =20$

$\Rightarrow \lambda =\frac{5}{4}$, hence option $\left(B\right)$ is correct and option $\left(C\right)$ is wrong,

Now solving option $\left(D\right)$ we get,

$\overrightarrow{b}\times \overrightarrow{d}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 3& 6& 3\\ 2& 1& 1\end{array}\right|=3\hat{i}+3\hat{j}-9\hat{k}$

$\Rightarrow {\left|\overrightarrow{b}\times \overrightarrow{d}\right|}^2={\left(\sqrt{3^2+3^2+9^2}\right)}^2=99$,

Hence, option $\left(D\right)$ is wrong.