If the vectors, $$\vec{p} = (a+1)\hat{i} + a\hat{j} + a\hat{k}$$, $$\vec{q} = a\hat{i} + (a+1)\hat{j} + a\hat{k}$$ and $$\vec{r} = a\hat{i} + a\hat{j} + (a+1)\hat{k}$$ $$(a \in R)$$ are coplanar and $$3(\vec{p} \cdot \vec{q})^2 - \lambda|\vec{r} \times \vec{q}|^2 = 0$$, then the value of $$\lambda$$ is ___________.

To ensure that the three given vectors are coplanar we recall the scalar-triple-product condition: for vectors $$\vec{p},\;\vec{q},\;\vec{r}$$ we must have $$\vec{p}\cdot(\vec{q}\times\vec{r})=0.$$

Writing each vector in component form, we obtain the determinant

$$$\vec{p}\cdot(\vec{q}\times\vec{r})= \begin{vmatrix} a+1 & a & a\\ a & a+1 & a\\ a & a & a+1 \end{vmatrix}.$$$

Expanding this determinant along the first row, we get

$$$\begin{aligned} \bigl(a+1\bigr)\Bigl[(a+1)(a+1)-a^{2}\Bigr] \;&-;; a\Bigl[a(a+1)-a^{2}\Bigr] \;+\; a\Bigl[a^{2}-a(a+1)\Bigr]. \end{aligned}$$$

Simplifying each bracket separately,

$$$\begin{aligned} (a+1)(a+1)-a^{2}&=a^{2}+2a+1-a^{2}=2a+1,\\[2mm] a(a+1)-a^{2}&=a^{2}+a-a^{2}=a,\\[2mm] a^{2}-a(a+1)&=a^{2}-a^{2}-a=-a. \end{aligned}$$$

Substituting these back,

$$$\begin{aligned} \vec{p}\cdot(\vec{q}\times\vec{r}) &=(a+1)(2a+1)-a(a)+a(-a)\\[2mm] &=2a^{2}+3a+1-a^{2}-a^{2}\\[2mm] &=3a+1. \end{aligned}$$$

The coplanarity requirement $$3a+1=0$$ therefore gives

$$a=-\dfrac13.$$

Next we evaluate $$\vec{p}\cdot\vec{q}.$$ Using the usual dot-product formula,

$$$\begin{aligned} \vec{p}\cdot\vec{q} &=(a+1)(a)+a(a+1)+a\cdot a\\[2mm] &=a(a+1)+a(a+1)+a^{2}\\[2mm] &=2a(a+1)+a^{2}\\[2mm] &=2(a^{2}+a)+a^{2}\\[2mm] &=3a^{2}+2a. \end{aligned}$$$

Putting $$a=-\dfrac13$$ we obtain

$$$\vec{p}\cdot\vec{q}=3\!\left(\dfrac{1}{9}\right)+2\!\left(-\dfrac13\right) =\dfrac13-\dfrac23=-\dfrac13.$$$

We now compute $$\vec{r}\times\vec{q}.$$ For components $$\vec{r}=(a,a,a+1)$$ and $$\vec{q}=(a,a+1,a),$$ the cross-product formula

$$$\vec{r}\times\vec{q}=\bigl(r_{2}q_{3}-r_{3}q_{2},\;r_{3}q_{1}-r_{1}q_{3},\;r_{1}q_{2}-r_{2}q_{1}\bigr)$$$

gives

$$$\begin{aligned} r_{2}q_{3}-r_{3}q_{2}&=a\cdot a-(a+1)(a+1)=a^{2}-(a+1)^{2}=-(2a+1),\\[2mm] r_{3}q_{1}-r_{1}q_{3}&=(a+1)\,a-a\cdot a=a(a+1)-a^{2}=a,\\[2mm] r_{1}q_{2}-r_{2}q_{1}&=a\,(a+1)-a\cdot a=a(a+1)-a^{2}=a. \end{aligned}$$$

Hence

$$\vec{r}\times\vec{q}=(-(2a+1),\,a,\,a).$$

The square of its magnitude is therefore

$$$\begin{aligned} \bigl|\vec{r}\times\vec{q}\bigr|^{2} &=(2a+1)^{2}+a^{2}+a^{2}\\[2mm] &=4a^{2}+4a+1+2a^{2}\\[2mm] &=6a^{2}+4a+1. \end{aligned}$$$

Substituting $$a=-\dfrac13,$$

$$$\bigl|\vec{r}\times\vec{q}\bigr|^{2}=6\!\left(\dfrac19\right)+4\!\left(-\dfrac13\right)+1 =\dfrac23-\dfrac43+1=\dfrac13.$$$

The problem states that

$$3(\vec{p}\cdot\vec{q})^{2}-\lambda\bigl|\vec{r}\times\vec{q}\bigr|^{2}=0.$$

Using $$(\vec{p}\cdot\vec{q})^{2}=\left(-\dfrac13\right)^{2}=\dfrac19$$ and $$\bigl|\vec{r}\times\vec{q}\bigr|^{2}=\dfrac13,$$ we have

$$$3\!\left(\dfrac19\right)-\lambda\!\left(\dfrac13\right)=0 \;\;\Longrightarrow\;\; \dfrac13-\dfrac{\lambda}{3}=0.$$$

Multiplying by 3 gives $$1-\lambda=0,$$ so

$$\lambda=1.$$

Hence, the correct answer is Option A.