If the vectors overrightarrow{p}=(a+1) hat{i} | vedclass

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(B) Given vectors are $\overrightarrow{p}=(a+1) \hat{i}+a \hat{j}+a \hat{k}$,$\overrightarrow{q}=a \hat{i}+(a+1) \hat{j}+a \hat{k}$,and $\overrightarr

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$1$

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(B) Given vectors are $\overrightarrow{p}=(a+1) \hat{i}+a \hat{j}+a \hat{k}$,$\overrightarrow{q}=a \hat{i}+(a+1) \hat{j}+a \hat{k}$,and $\overrightarrow{r}=a \hat{i}+a \hat{j}+(a+1) \hat{k}$.Since $\overrightarrow{p}, \overrightarrow{q}, \overrightarrow{r}$ are coplanar,their scalar triple product is zero:$\left|\begin{array}{ccc} a+1 & a & a \ a & a+1 & a \ a & a & a+1 \end{array}ight|=0$Applying $R_1 	o R_1+R_2+R_3$,we get $(3a+1) \left|\begin{array}{ccc} 1 & 1 & 1 \ a & a+1 & a \ a & a & a+1 \end{array}ight|=0$,which simplifies to $(3a+1)(1)=0$,so $a = -\frac{1}{3}$.Now,$\overrightarrow{p} \cdot \overrightarrow{q} = a^2 + a(a+1) + a^2 = 3a^2+a = 3(-\frac{1}{3})^2 + (-\frac{1}{3}) = \frac{1}{3} - \frac{1}{3} = 0$. Wait,let's recompute: $\overrightarrow{p} \cdot \overrightarrow{q} = a(a+1) + a(a+1) + a^2 = 2a^2+2a+a^2 = 3a^2+2a$. For $a=-1/3$,$\overrightarrow{p} \cdot \overrightarrow{q} = 3(1/9) - 2/3 = 1/3 - 2/3 = -1/3$.Also,$|\overrightarrow{r}|^2 = |\overrightarrow{q}|^2 = a^2 + a^2 + (a+1)^2 = 3a^2+2a+1 = 3(1/9) - 2/3 + 1 = 1/3 - 2/3 + 1 = 2/3$.$\overrightarrow{r} \cdot \overrightarrow{q} = a^2 + a(a+1) + a(a+1) = a^2 + 2a^2 + 2a = 3a^2+2a = -1/3$.Using Lagrange's identity,$|\overrightarrow{r} 	imes \overrightarrow{q}|^2 = |\overrightarrow{r}|^2 |\overrightarrow{q}|^2 - (\overrightarrow{r} \cdot \overrightarrow{q})^2 = (2/3)(2/3) - (-1/3)^2 = 4/9 - 1/9 = 3/9 = 1/3$.Given $3(\overrightarrow{p} \cdot \overrightarrow{q})^2 - \lambda |\overrightarrow{r} 	imes \overrightarrow{q}|^2 = 0$,we have $3(-1/3)^2 - \lambda(1/3) = 0 \Rightarrow 3(1/9) - \lambda/3 = 0 \Rightarrow 1/3 = \lambda/3 \Rightarrow \lambda = 1$.

Column $I$	Column $II$
$(A)$ Root$(s)$ of the equation $2 \sin ^2 \theta + \sin ^2 2 \theta = 2$	$(p)$ $\frac{\pi}{6}$
$(B)$ Points of discontinuity of the function $f(x) = [\frac{6x}{\pi}] \cos [\frac{3x}{\pi}]$,where $[y]$ denotes the largest integer less than or equal to $y$	$(q)$ $\frac{\pi}{4}$
$(C)$ Volume of the parallelepiped with its edges represented by the vectors $\hat{i}+\hat{j}, \hat{i}+2\hat{j}$ and $\hat{i}+\hat{j}+\pi\hat{k}$	$(r)$ $\frac{\pi}{3}$
$(D)$ Angle between vectors $\vec{a}$ and $\vec{b}$ where $\vec{a}, \vec{b}$ and $\vec{c}$ are unit vectors satisfying $\vec{a}+\vec{b}+\sqrt{3}\vec{c}=\overrightarrow{0}$	$(s)$ $\frac{\pi}{2}$
	$(t)$ $\pi$

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