Let the position vectors of points P, Q, R and S be → a = ^ i + 2 ^...

Given,

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

Now solving options,

Option \left(A\right)$checking points are coplanar,</p><p>$\overset{\to}{PQ} = \left(2, 4, 8\right), \overset{\to}{PR} = \left(\dfrac{12}{5}, \dfrac{6}{5}, 12\right) \text{$and$} \overset{\to}{PQ} = \left(1, - 1, 6\right)</p><p>Now if points are coplanar then, \left[\begin{cases}\overset{\to}{PQ} & \overset{\to}{PR} & \overset{\to}{PS}\end{cases}\right] = 0

Now solving L.H.S, we get,

\left|$\begin{cases}2 & 4 & 8 \\ \dfrac{12}{5} & \dfrac{6}{5} & 12 \\ 1 & - 1 & 6\end{cases}$\right| = \dfrac{2}{5}\left|$\begin{cases}1 & 2 & 4 \\ 12 & 6 & 60 \\ 1 & - 1 & 6\end{cases}$\right| = \dfrac{2}{5}\left(96 - 24 - 72\right) = 0$</p><p>Hence, they are coplanar</p><p>So option \left(A\right)$ is wrong.

Now solving, option \left(B\right)$we get,</p><p>$\dfrac{\overset{\to}{b} + 2\overset{\to}{d}}{3} = \dfrac{7i + 8j + 5k}{3}$</p><p>Now let point divides the$PR\text{in}\lambda: 1$we get,</p><p>So, on taking coefficient of$\overset{^}{i}$we get,</p><p>$⇒ \dfrac{\dfrac{17\lambda}{5} + 1}{( \lambda + 1 )} = \dfrac{7}{3}$</p><p>$⇒ \dfrac{17\lambda}{5} + 1 = \dfrac{7}{3} ( \lambda + 1 )$</p><p>$⇒ 51\lambda + 15 = 35\lambda + 35$</p><p>$⇒ 16\lambda = 20$</p><p>$⇒ \lambda = \dfrac{5}{4}$, hence option \left(B\right)$ is correct and option \left(C\right)$is wrong,</p><p>Now solving option \left(D\right)$ we get,

\overset{\to}{b} \times \overset{\to}{d} = \left|\begin{cases}\overset{^}{i} & \overset{^}{j} & \overset{^}{k} \\ 3 & 6 & 3 \\ 2 & 1 & 1\end{cases}\right| = 3\overset{^}{i} + 3\overset{^}{j} - 9\overset{^}{k}

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

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