Scalar triple product and their applications Questions in English

(D) The volume of a parallelepiped with coterminous edges $\vec{u}, \vec{v}, \vec{w}$ is given by the scalar triple product $[\vec{u} \vec{v} \vec{w}] = \vec{u} \cdot (\vec{v} \times \vec{w})$.
Here,$\vec{u} = \bar{a}+\bar{b}$,$\vec{v} = \bar{b}+\bar{c}$,and $\vec{w} = \bar{c}+\bar{a}$.
Volume $= (\bar{a}+\bar{b}) \cdot [(\bar{b}+\bar{c}) \times (\bar{c}+\bar{a})]$
$= (\bar{a}+\bar{b}) \cdot [(\bar{b} \times \bar{c}) + (\bar{b} \times \bar{a}) + (\bar{c} \times \bar{c}) + (\bar{c} \times \bar{a})]$
Since $\bar{c} \times \bar{c} = 0$,we have:
$= (\bar{a}+\bar{b}) \cdot [(\bar{b} \times \bar{c}) + (\bar{b} \times \bar{a}) + (\bar{c} \times \bar{a})]$
$= \bar{a} \cdot (\bar{b} \times \bar{c}) + \bar{a} \cdot (\bar{b} \times \bar{a}) + \bar{a} \cdot (\bar{c} \times \bar{a}) + \bar{b} \cdot (\bar{b} \times \bar{c}) + \bar{b} \cdot (\bar{b} \times \bar{a}) + \bar{b} \cdot (\bar{c} \times \bar{a})$
Using the property that the scalar triple product is zero if any two vectors are identical:
$= [\bar{a} \bar{b} \bar{c}] + 0 + 0 + 0 + 0 + [\bar{b} \bar{c} \bar{a}]$
Since $[\bar{b} \bar{c} \bar{a}] = [\bar{a} \bar{b} \bar{c}]$,the volume is $[\bar{a} \bar{b} \bar{c}] + [\bar{a} \bar{b} \bar{c}] = 2[\bar{a} \bar{b} \bar{c}]$.

(C) Let the given vectors be $\vec{a} = \hat{\imath} - \hat{\jmath} - 6\hat{k}$,$\vec{b} = \hat{\imath} - 3\hat{\jmath} + 4\hat{k}$,and $\vec{c} = 2\hat{\imath} - 5\hat{\jmath} + m\hat{k}$.
Since the vectors are coplanar,their scalar triple product must be zero,i.e.,$[\vec{a} \vec{b} \vec{c}] = 0$.
This implies the determinant of the components is zero:
$\begin{vmatrix} 1 & -1 & -6 \\ 1 & -3 & 4 \\ 2 & -5 & m \end{vmatrix} = 0$
Expanding along the first row:
$1((-3)(m) - (4)(-5)) - (-1)((1)(m) - (4)(2)) + (-6)((1)(-5) - (-3)(2)) = 0$
$1(-3m + 20) + 1(m - 8) - 6(-5 + 6) = 0$
$-3m + 20 + m - 8 - 6(1) = 0$
$-2m + 12 - 6 = 0$
$-2m + 6 = 0$
$2m = 6$
$m = 3$

(A) The volume of a parallelepiped with coterminous edges $\vec{u}, \vec{v}, \vec{w}$ is given by the scalar triple product $[\vec{u} \quad \vec{v} \quad \vec{w}] = \vec{u} \cdot (\vec{v} \times \vec{w})$.
Given $[\bar{a} \quad \bar{b} \quad \bar{c}] = 4$.
We need to find the volume of the parallelepiped with edges $\bar{a}+2 \bar{b}$,$\bar{b}+2 \bar{c}$,and $\bar{c}+2 \bar{a}$.
Volume $= [(\bar{a}+2 \bar{b}) \quad (\bar{b}+2 \bar{c}) \quad (\bar{c}+2 \bar{a})]$.
$= (\bar{a}+2 \bar{b}) \cdot [(\bar{b}+2 \bar{c}) \times (\bar{c}+2 \bar{a})]$.
$= (\bar{a}+2 \bar{b}) \cdot [(\bar{b} \times \bar{c}) + 2(\bar{b} \times \bar{a}) + 2(\bar{c} \times \bar{c}) + 4(\bar{c} \times \bar{a})]$.
Since $\bar{c} \times \bar{c} = 0$,this simplifies to:
$= (\bar{a}+2 \bar{b}) \cdot [(\bar{b} \times \bar{c}) + 2(\bar{b} \times \bar{a}) + 4(\bar{c} \times \bar{a})]$.
$= \bar{a} \cdot (\bar{b} \times \bar{c}) + 2\bar{a} \cdot (\bar{b} \times \bar{a}) + 4\bar{a} \cdot (\bar{c} \times \bar{a}) + 2\bar{b} \cdot (\bar{b} \times \bar{c}) + 4\bar{b} \cdot (\bar{b} \times \bar{a}) + 8\bar{b} \cdot (\bar{c} \times \bar{a})$.
Using the property that the scalar triple product is zero if any two vectors are the same:
$= [\bar{a} \quad \bar{b} \quad \bar{c}] + 0 + 0 + 0 + 0 + 8[\bar{b} \quad \bar{c} \quad \bar{a}]$.
Since $[\bar{b} \quad \bar{c} \quad \bar{a}] = [\bar{a} \quad \bar{b} \quad \bar{c}] = 4$,
Volume $= 4 + 8(4) = 4 + 32 = 36 \text{ units}^3$.

(C) First,calculate the cross product $\bar{u} \times \bar{v}$:
$\bar{u} \times \bar{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 1 \\ 3 & 0 & 1 \end{vmatrix} = \hat{i}(-2-0) - \hat{j}(1-3) + \hat{k}(0 - (-6)) = -2\hat{i} + 2\hat{j} + 6\hat{k}$.
Next,define the vectors for the edges:
$\bar{a} = \bar{u} \times \bar{v} = -2\hat{i} + 2\hat{j} + 6\hat{k}$
$\bar{b} = \bar{u} + \bar{w} = (\hat{i} - 2\hat{j} + \hat{k}) + (\hat{j} - \hat{k}) = \hat{i} - \hat{j}$
$\bar{c} = \bar{v} + \bar{w} = (3\hat{i} + \hat{k}) + (\hat{j} - \hat{k}) = 3\hat{i} + \hat{j}$
The volume of the parallelepiped is given by the scalar triple product $|\bar{a} \cdot (\bar{b} \times \bar{c})|$,which is equivalent to the absolute value of the determinant of the matrix formed by these vectors:
Volume $= \left| \begin{vmatrix} -2 & 2 & 6 \\ 1 & -1 & 0 \\ 3 & 1 & 0 \end{vmatrix} \right|$
Expanding along the third column:
Volume $= |6(1 - (-3))| = |6(4)| = 24$ cubic units.

(A) The volume of a tetrahedron with vertices $A, B, C, D$ is given by the formula $V = \frac{1}{6} |[\vec{AB} \vec{AC} \vec{AD}]|$.
First,we find the vectors:
$\vec{AB} = (3 - (-1))\hat{i} + (-2 - 2)\hat{j} + (1 - 3)\hat{k} = 4\hat{i} - 4\hat{j} - 2\hat{k}$
$\vec{AC} = (2 - (-1))\hat{i} + (1 - 2)\hat{j} + (3 - 3)\hat{k} = 3\hat{i} - 1\hat{j} + 0\hat{k}$
$\vec{AD} = (-1 - (-1))\hat{i} + (-2 - 2)\hat{j} + (4 - 3)\hat{k} = 0\hat{i} - 4\hat{j} + 1\hat{k}$
Now,calculate the scalar triple product $[\vec{AB} \vec{AC} \vec{AD}]$ using the determinant:
$|\vec{AB} \vec{AC} \vec{AD}| = \begin{vmatrix} 4 & -4 & -2 \\ 3 & -1 & 0 \\ 0 & -4 & 1 \end{vmatrix}$
$= 4((-1)(1) - (0)(-4)) - (-4)((3)(1) - (0)(0)) + (-2)((3)(-4) - (-1)(0))$
$= 4(-1) + 4(3) - 2(-12) = -4 + 12 + 24 = 32$
Therefore,the volume $V = \frac{1}{6} \times 32 = \frac{32}{6} = \frac{16}{3}$ cu. units.

(C) The scalar triple product is defined as $[\bar{a} \quad \bar{b} \quad \bar{c}] = \bar{a} \cdot (\bar{b} \times \bar{c})$.
We need to evaluate the expression $[\bar{a}+2\bar{b} \quad \bar{b}+2\bar{c} \quad \bar{c}+2\bar{a}]$.
Using the properties of the scalar triple product,we can write this as:
$[\bar{a}+2\bar{b} \quad \bar{b}+2\bar{c} \quad \bar{c}+2\bar{a}] = (\bar{a}+2\bar{b}) \cdot [(\bar{b}+2\bar{c}) \times (\bar{c}+2\bar{a})]$.
Expanding the cross product:
$(\bar{b}+2\bar{c}) \times (\bar{c}+2\bar{a}) = (\bar{b} \times \bar{c}) + 2(\bar{b} \times \bar{a}) + 2(\bar{c} \times \bar{c}) + 4(\bar{c} \times \bar{a})$.
Since $\bar{c} \times \bar{c} = 0$,this simplifies to $(\bar{b} \times \bar{c}) + 2(\bar{b} \times \bar{a}) + 4(\bar{c} \times \bar{a})$.
Now,take the dot product with $(\bar{a}+2\bar{b})$:
$(\bar{a}+2\bar{b}) \cdot [(\bar{b} \times \bar{c}) + 2(\bar{b} \times \bar{a}) + 4(\bar{c} \times \bar{a})]$
$= \bar{a} \cdot (\bar{b} \times \bar{c}) + 2\bar{a} \cdot (\bar{b} \times \bar{a}) + 4\bar{a} \cdot (\bar{c} \times \bar{a}) + 2\bar{b} \cdot (\bar{b} \times \bar{c}) + 4\bar{b} \cdot (\bar{b} \times \bar{a}) + 8\bar{b} \cdot (\bar{c} \times \bar{a})$.
Using the property that the scalar triple product is zero if any two vectors are identical:
$= [\bar{a} \bar{b} \bar{c}] + 0 + 0 + 0 + 0 + 8[\bar{b} \bar{c} \bar{a}]$
$= [\bar{a} \bar{b} \bar{c}] + 8[\bar{a} \bar{b} \bar{c}] = 9[\bar{a} \bar{b} \bar{c}]$.
Therefore,$\frac{[\bar{a}+2\bar{b} \quad \bar{b}+2\bar{c} \quad \bar{c}+2\bar{a}]}{[\bar{a} \bar{b} \bar{c}]} = \frac{9[\bar{a} \bar{b} \bar{c}]}{[\bar{a} \bar{b} \bar{c}]} = 9$.

(C) The volume of a parallelepiped with coterminus edges $\vec{u}, \vec{v}, \vec{w}$ is given by the scalar triple product $[\vec{u} \vec{v} \vec{w}] = \vec{u} \cdot (\vec{v} \times \vec{w})$.
Given $\vec{u} = 2\bar{a}+\bar{b}$,$\vec{v} = 2\bar{b}+\vec{c}$,and $\vec{w} = 2\bar{c}+\bar{a}$.
Volume $= (2\bar{a}+\bar{b}) \cdot [(2\bar{b}+\bar{c}) \times (2\bar{c}+\bar{a})]$.
Expanding the cross product: $(2\bar{b}+\bar{c}) \times (2\bar{c}+\bar{a}) = 4(\bar{b} \times \bar{c}) + 2(\bar{b} \times \bar{a}) + 2(\bar{c} \times \bar{c}) + (\bar{c} \times \bar{a}) = 4(\bar{b} \times \bar{c}) + 2(\bar{b} \times \bar{a}) + (\bar{c} \times \bar{a})$.
Now,taking the dot product with $(2\bar{a}+\bar{b})$:
Volume $= (2\bar{a}+\bar{b}) \cdot [4(\bar{b} \times \bar{c}) + 2(\bar{b} \times \bar{a}) + (\bar{c} \times \bar{a})]$.
$= 8[\bar{a} \bar{b} \bar{c}] + 4[\bar{a} \bar{b} \bar{a}] + 2[\bar{a} \bar{c} \bar{a}] + 4[\bar{b} \bar{b} \bar{c}] + 2[\bar{b} \bar{b} \bar{a}] + [\bar{b} \bar{c} \bar{a}]$.
Since the scalar triple product is zero if any two vectors are identical,we have $[\bar{a} \bar{b} \bar{a}] = 0, [\bar{a} \bar{c} \bar{a}] = 0, [\bar{b} \bar{b} \bar{c}] = 0, [\bar{b} \bar{b} \bar{a}] = 0$.
Volume $= 8[\bar{a} \bar{b} \bar{c}] + [\bar{b} \bar{c} \bar{a}] = 8[\bar{a} \bar{b} \bar{c}] + [\bar{a} \bar{b} \bar{c}] = 9[\bar{a} \bar{b} \bar{c}]$.
Given $[\bar{a} \bar{b} \bar{c}] = 3$,Volume $= 9 \times 3 = 27$ cubic units.

(B) The volume of a tetrahedron with vertices $A, B, C, D$ is given by $V = \frac{1}{6} |(\vec{AB}) \cdot (\vec{AC} \times \vec{AD})|$.
First,we find the vectors:
$\vec{AB} = (3 - (-1))\hat{i} + (-2 - 2)\hat{j} + (1 - 3)\hat{k} = 4\hat{i} - 4\hat{j} - 2\hat{k}$
$\vec{AC} = (2 - (-1))\hat{i} + (1 - 2)\hat{j} + (3 - 3)\hat{k} = 3\hat{i} - 1\hat{j} + 0\hat{k}$
$\vec{AD} = (-1 - (-1))\hat{i} + (-2 - 2)\hat{j} + (4 - 3)\hat{k} = 0\hat{i} - 4\hat{j} + 1\hat{k}$
Now,calculate the scalar triple product:
$V = \frac{1}{6} \left| \begin{vmatrix} 4 & -4 & -2 \\ 3 & -1 & 0 \\ 0 & -4 & 1 \end{vmatrix} \right|$
$V = \frac{1}{6} |4(-1 - 0) - (-4)(3 - 0) + (-2)(-12 - 0)|$
$V = \frac{1}{6} |4(-1) + 4(3) - 2(-12)|$
$V = \frac{1}{6} |-4 + 12 + 24| = \frac{1}{6} |32| = \frac{32}{6} = \frac{16}{3} \text{ cu. units}$.

(C) We know that the scalar triple product $[\bar{a}+\bar{b} \quad \bar{b}+\bar{c} \quad \bar{c}+\bar{a}]$ can be expanded as follows:
$[\bar{a}+\bar{b} \quad \bar{b}+\bar{c} \quad \bar{c}+\bar{a}] = (\bar{a}+\bar{b}) \cdot [(\bar{b}+\bar{c}) \times (\bar{c}+\bar{a})]$
$= (\bar{a}+\bar{b}) \cdot [(\bar{b} \times \bar{c}) + (\bar{b} \times \bar{a}) + (\bar{c} \times \bar{c}) + (\bar{c} \times \bar{a})]$
Since $\bar{c} \times \bar{c} = 0$,we have:
$= (\bar{a}+\bar{b}) \cdot [(\bar{b} \times \bar{c}) + (\bar{b} \times \bar{a}) + (\bar{c} \times \bar{a})]$
$= \bar{a} \cdot (\bar{b} \times \bar{c}) + \bar{a} \cdot (\bar{b} \times \bar{a}) + \bar{a} \cdot (\bar{c} \times \bar{a}) + \bar{b} \cdot (\bar{b} \times \bar{c}) + \bar{b} \cdot (\bar{b} \times \bar{a}) + \bar{b} \cdot (\bar{c} \times \bar{a})$
Using the property that the scalar triple product is zero if any two vectors are identical,we get:
$= [\bar{a} \bar{b} \bar{c}] + 0 + 0 + 0 + 0 + [\bar{b} \bar{c} \bar{a}]$
Since $[\bar{a} \bar{b} \bar{c}] = [\bar{b} \bar{c} \bar{a}]$,we have:
$= [\bar{a} \bar{b} \bar{c}] + [\bar{a} \bar{b} \bar{c}] = 2[\bar{a} \bar{b} \bar{c}]$
Therefore,the given expression is $\frac{2[\bar{a} \bar{b} \bar{c}]}{[\bar{a} \bar{b} \bar{c}]} = 2$.

(C) For three vectors to be coplanar,their scalar triple product must be zero.
Let the vectors be $\vec{a} = \hat{\imath}+\hat{\jmath}+\hat{k}$,$\vec{b} = \hat{\imath}-\hat{\jmath}+\hat{k}$,and $\vec{c} = 2\hat{\imath}+3\hat{\jmath}+m\hat{k}$.
The condition for coplanarity is $\vec{a} \cdot (\vec{b} \times \vec{c}) = 0$,which is equivalent to the determinant of the components being zero:
$\begin{vmatrix} 1 & 1 & 1 \\ 1 & -1 & 1 \\ 2 & 3 & m \end{vmatrix} = 0$
Expanding the determinant along the first row:
$1((-1)(m) - (1)(3)) - 1((1)(m) - (1)(2)) + 1((1)(3) - (-1)(2)) = 0$
$1(-m - 3) - 1(m - 2) + 1(3 + 2) = 0$
$-m - 3 - m + 2 + 5 = 0$
$-2m + 4 = 0$
$2m = 4$
$m = 2$

(A) Given that the volume of the parallelepiped formed by vectors $\bar{a}, \bar{b}, \bar{c}$ is $12$. Thus,the scalar triple product is $[\bar{a} \bar{b} \bar{c}] = 12$.
The volume of a tetrahedron with coterminous edges $\bar{u}, \bar{v}, \bar{w}$ is given by $\frac{1}{6} |[\bar{u} \bar{v} \bar{w}]|$.
Here,the edges are $\bar{a}+\bar{b}, \bar{b}+\bar{c}, \bar{c}+\bar{a}$.
We calculate the scalar triple product:
$[\bar{a}+\bar{b} \quad \bar{b}+\bar{c} \quad \bar{c}+\bar{a}] = (\bar{a}+\bar{b}) \cdot ((\bar{b}+\bar{c}) \times (\bar{c}+\bar{a}))$
$= (\bar{a}+\bar{b}) \cdot (\bar{b} \times \bar{c} + \bar{b} \times \bar{a} + \bar{c} \times \bar{c} + \bar{c} \times \bar{a})$
Since $\bar{c} \times \bar{c} = 0$,this simplifies to:
$= (\bar{a}+\bar{b}) \cdot (\bar{b} \times \bar{c} + \bar{b} \times \bar{a} + \bar{c} \times \bar{a})$
$= \bar{a} \cdot (\bar{b} \times \bar{c}) + \bar{a} \cdot (\bar{b} \times \bar{a}) + \bar{a} \cdot (\bar{c} \times \bar{a}) + \bar{b} \cdot (\bar{b} \times \bar{c}) + \bar{b} \cdot (\bar{b} \times \bar{a}) + \bar{b} \cdot (\bar{c} \times \bar{a})$
$= [\bar{a} \bar{b} \bar{c}] + 0 + 0 + 0 + 0 + [\bar{b} \bar{c} \bar{a}]$
$= [\bar{a} \bar{b} \bar{c}] + [\bar{a} \bar{b} \bar{c}] = 2[\bar{a} \bar{b} \bar{c}] = 2(12) = 24$.
Therefore,the volume of the tetrahedron is $\frac{1}{6} \times 24 = 4 \text{ (units)}^3$.

(B) The scalar triple product is defined as $[\bar{b}, \bar{a} \times \bar{b}, \bar{a}] = \bar{b} \cdot ((\bar{a} \times \bar{b}) \times \bar{a})$.
Using the property of the vector triple product,$(\bar{a} \times \bar{b}) \times \bar{a} = -\bar{a} \times (\bar{a} \times \bar{b})$.
By the vector triple product formula $\bar{u} \times (\bar{v} \times \bar{w}) = (\bar{u} \cdot \bar{w})\bar{v} - (\bar{u} \cdot \bar{v})\bar{w}$,we have $\bar{a} \times (\bar{a} \times \bar{b}) = (\bar{a} \cdot \bar{b})\bar{a} - (\bar{a} \cdot \bar{a})\bar{b}$.
Thus,$[\bar{b}, \bar{a} \times \bar{b}, \bar{a}] = \bar{b} \cdot ((\bar{a} \cdot \bar{b})\bar{a} - (\bar{a} \cdot \bar{a})\bar{b}) = (\bar{a} \cdot \bar{b})(\bar{b} \cdot \bar{a}) - (\bar{a} \cdot \bar{a})(\bar{b} \cdot \bar{b}) = (\bar{a} \cdot \bar{b})^2 - |\bar{a}|^2 |\bar{b}|^2$.
Using Lagrange's identity,$|\bar{a} \times \bar{b}|^2 = |\bar{a}|^2 |\bar{b}|^2 - (\bar{a} \cdot \bar{b})^2$.
Therefore,$[\bar{b}, \bar{a} \times \bar{b}, \bar{a}] = -|\bar{a} \times \bar{b}|^2$.

(C) The volume of a parallelepiped with coterminus edges $\bar{a}, \bar{b}, \bar{c}$ is given by the scalar triple product $[\bar{a} \bar{b} \bar{c}] = |\bar{a} \cdot (\bar{b} \times \bar{c})| = 7$.
We need to find the volume of a parallelepiped with edges $\bar{a}+\bar{b}, \bar{b}+\bar{c}, \bar{c}+\bar{a}$.
This volume is given by the scalar triple product $[\bar{a}+\bar{b} \quad \bar{b}+\bar{c} \quad \bar{c}+\bar{a}]$.
Using the property of the scalar triple product,we have:
$[\bar{a}+\bar{b} \quad \bar{b}+\bar{c} \quad \bar{c}+\bar{a}] = (\bar{a}+\bar{b}) \cdot ((\bar{b}+\bar{c}) \times (\bar{c}+\bar{a}))$.
Expanding the cross product:
$(\bar{b}+\bar{c}) \times (\bar{c}+\bar{a}) = \bar{b} \times \bar{c} + \bar{b} \times \bar{a} + \bar{c} \times \bar{c} + \bar{c} \times \bar{a} = \bar{b} \times \bar{c} + \bar{b} \times \bar{a} + \bar{c} \times \bar{a}$ (since $\bar{c} \times \bar{c} = 0$).
Now,taking the dot product with $(\bar{a}+\bar{b})$:
$(\bar{a}+\bar{b}) \cdot (\bar{b} \times \bar{c} + \bar{b} \times \bar{a} + \bar{c} \times \bar{a}) = \bar{a} \cdot (\bar{b} \times \bar{c}) + \bar{a} \cdot (\bar{b} \times \bar{a}) + \bar{a} \cdot (\bar{c} \times \bar{a}) + \bar{b} \cdot (\bar{b} \times \bar{c}) + \bar{b} \cdot (\bar{b} \times \bar{a}) + \bar{b} \cdot (\bar{c} \times \bar{a})$.
Since the scalar triple product is zero if any two vectors are the same,we have:
$= [\bar{a} \bar{b} \bar{c}] + 0 + 0 + 0 + 0 + [\bar{b} \bar{c} \bar{a}] = [\bar{a} \bar{b} \bar{c}] + [\bar{a} \bar{b} \bar{c}] = 2[\bar{a} \bar{b} \bar{c}]$.
Given $[\bar{a} \bar{b} \bar{c}] = 7$,the volume is $2 \times 7 = 14$ cubic units.

(C) Given that $\bar{p}=\frac{\bar{b} \times \bar{c}}{[\bar{a} \bar{b} \bar{c}]}, \bar{q}=\frac{\bar{c} \times \bar{a}}{[\bar{a} \bar{b} \bar{c}]}, \bar{r}=\frac{\bar{a} \times \bar{b}}{[\bar{a} \bar{b} \bar{c}]}$.
We know that the scalar triple product $[\bar{a} \bar{b} \bar{c}] = \bar{a} \cdot (\bar{b} \times \bar{c})$.
Now,calculate each term:
$\bar{a} \cdot \bar{p} = \bar{a} \cdot \left( \frac{\bar{b} \times \bar{c}}{[\bar{a} \bar{b} \bar{c}]} \right) = \frac{\bar{a} \cdot (\bar{b} \times \bar{c})}{[\bar{a} \bar{b} \bar{c}]} = \frac{[\bar{a} \bar{b} \bar{c}]}{[\bar{a} \bar{b} \bar{c}]} = 1$.
Similarly,$\bar{b} \cdot \bar{q} = \bar{b} \cdot \left( \frac{\bar{c} \times \bar{a}}{[\bar{a} \bar{b} \bar{c}]} \right) = \frac{\bar{b} \cdot (\bar{c} \times \bar{a})}{[\bar{a} \bar{b} \bar{c}]} = \frac{[\bar{b} \bar{c} \bar{a}]}{[\bar{a} \bar{b} \bar{c}]} = \frac{[\bar{a} \bar{b} \bar{c}]}{[\bar{a} \bar{b} \bar{c}]} = 1$.
And $\bar{c} \cdot \bar{r} = \bar{c} \cdot \left( \frac{\bar{a} \times \bar{b}}{[\bar{a} \bar{b} \bar{c}]} \right) = \frac{\bar{c} \cdot (\bar{a} \times \bar{b})}{[\bar{a} \bar{b} \bar{c}]} = \frac{[\bar{c} \bar{a} \bar{b}]}{[\bar{a} \bar{b} \bar{c}]} = \frac{[\bar{a} \bar{b} \bar{c}]}{[\bar{a} \bar{b} \bar{c}]} = 1$.
Therefore,$\bar{a} \cdot \bar{p} + \bar{b} \cdot \bar{q} + \bar{c} \cdot \bar{r} = 1 + 1 + 1 = 3$.

(D) We expand the dot product: $(\bar{a} + \bar{b} + \bar{c}) \cdot (\bar{a} \times \bar{b} + \bar{b} \times \bar{c} + \bar{c} \times \bar{a})$.
Using the distributive property of the dot product,we get:
$= \bar{a} \cdot (\bar{a} \times \bar{b}) + \bar{a} \cdot (\bar{b} \times \bar{c}) + \bar{a} \cdot (\bar{c} \times \bar{a}) + \bar{b} \cdot (\bar{a} \times \bar{b}) + \bar{b} \cdot (\bar{b} \times \bar{c}) + \bar{b} \cdot (\bar{c} \times \bar{a}) + \bar{c} \cdot (\bar{a} \times \bar{b}) + \bar{c} \cdot (\bar{b} \times \bar{c}) + \bar{c} \cdot (\bar{c} \times \bar{a})$.
Since the scalar triple product $[\bar{x} \bar{y} \bar{z}] = 0$ if any two vectors are identical,terms like $\bar{a} \cdot (\bar{a} \times \bar{b}) = 0$,$\bar{b} \cdot (\bar{b} \times \bar{c}) = 0$,and $\bar{c} \cdot (\bar{c} \times \bar{a}) = 0$.
This leaves us with:
$= [\bar{a} \bar{b} \bar{c}] + [\bar{b} \bar{c} \bar{a}] + [\bar{c} \bar{a} \bar{b}]$.
Since the scalar triple product is invariant under cyclic permutation,$[\bar{a} \bar{b} \bar{c}] = [\bar{b} \bar{c} \bar{a}] = [\bar{c} \bar{a} \bar{b}]$.
Therefore,the expression equals $3[\bar{a} \bar{b} \bar{c}]$.
Comparing this with $k[\bar{a} \bar{b} \bar{c}]$,we find $k = 3$.

(A) We have the expression $a \cdot[(b+c) \times(a+b+c)]$.
Using the distributive property of the cross product,we expand the term inside the bracket:
$(b+c) \times(a+b+c) = (b \times a) + (b \times b) + (b \times c) + (c \times a) + (c \times b) + (c \times c)$.
Since the cross product of any vector with itself is zero ($b \times b = 0$ and $c \times c = 0$) and $c \times b = -(b \times c)$,the expression simplifies to:
$(b \times a) + (b \times c) + (c \times a) - (b \times c) = (b \times a) + (c \times a)$.
Now,taking the dot product with $a$:
$a \cdot [(b \times a) + (c \times a)] = a \cdot (b \times a) + a \cdot (c \times a)$.
By the definition of the scalar triple product,$a \cdot (b \times a) = [a \ b \ a]$ and $a \cdot (c \times a) = [a \ c \ a]$.
Since any scalar triple product with two identical vectors is zero,we have $[a \ b \ a] = 0$ and $[a \ c \ a] = 0$.
Therefore,the final result is $0 + 0 = 0$.

(B) The scalar triple product of vectors $\vec{a}, \vec{b}, \vec{c}$ is given by the determinant of the matrix formed by their components: $[\vec{a} \vec{b} \vec{c}] = \begin{vmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{vmatrix}$.
Given the vectors $-3 \hat{i}+7 \hat{j}-3 \hat{k}$,$3 \hat{i}-7 \hat{j}+\lambda \hat{k}$,and $7 \hat{i}-5 \hat{j}-3 \hat{k}$,their scalar triple product is $272$.
$\begin{vmatrix} -3 & 7 & -3 \\ 3 & -7 & \lambda \\ 7 & -5 & -3 \end{vmatrix} = 272$
Expanding the determinant along the first row:
$-3((-7)(-3) - (-5)(\lambda)) - 7((3)(-3) - (7)(\lambda)) - 3((3)(-5) - (7)(-7)) = 272$
$-3(21 + 5\lambda) - 7(-9 - 7\lambda) - 3(-15 + 49) = 272$
$-63 - 15\lambda + 63 + 49\lambda - 3(34) = 272$
$34\lambda - 102 = 272$
$34\lambda = 374$
$\lambda = \frac{374}{34} = 11$.

(C) We know that the scalar triple product of unit vectors is given by the cyclic order property: $i \cdot(j \times k) = 1$,$j \cdot(k \times i) = 1$,and $k \cdot(i \times j) = 1$.
Given expression is $i \cdot(j \times k) + j \cdot(k \times i) + k \cdot(j \times i)$.
Since $j \times i = -k$,we have $k \cdot(j \times i) = k \cdot(-k) = -(k \cdot k) = -1$.
Substituting these values:
$i \cdot(j \times k) + j \cdot(k \times i) + k \cdot(j \times i) = 1 + 1 + (-1) = 1$.

(A) The position vectors of the vertices $A, B$ and $C$ with respect to the origin $O$ are given as:
$\vec{OA} = 6i = 6\hat{i} + 0\hat{j} + 0\hat{k}$
$\vec{OB} = 6j = 0\hat{i} + 6\hat{j} + 0\hat{k}$
$\vec{OC} = k = 0\hat{i} + 0\hat{j} + 1\hat{k}$
The volume of a tetrahedron with vertices at the origin and position vectors $\vec{a}, \vec{b}, \vec{c}$ is given by the formula:
$V = \frac{1}{6} |[\vec{a} \vec{b} \vec{c}]|$
Substituting the given vectors:
$V = \frac{1}{6} \left| \begin{vmatrix} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 1 \end{vmatrix} \right|$
Calculating the determinant:
$V = \frac{1}{6} (6 \times (6 \times 1 - 0 \times 0) - 0 + 0) = \frac{1}{6} (36) = 6$
Thus,the volume of the tetrahedron $OABC$ is $6$ cubic units.

(C) The volume of a parallelepiped with coterminous edges $\vec{u}, \vec{v}, \vec{w}$ is given by the scalar triple product $[\vec{u} \vec{v} \vec{w}] = (\vec{u} \times \vec{v}) \cdot \vec{w}$.
Given edges are $2\vec{a}, 2\vec{b}, 2\vec{c}$.
Volume $= [2\vec{a} \ 2\vec{b} \ 2\vec{c}]$
$= (2\vec{a} \times 2\vec{b}) \cdot 2\vec{c}$
$= 4(\vec{a} \times \vec{b}) \cdot 2\vec{c}$
$= 8(\vec{a} \times \vec{b}) \cdot \vec{c}$
$= 8[\vec{a} \vec{b} \vec{c}]$.

(D) Given that $\overrightarrow{p}=\frac{\overrightarrow{b} \times \overrightarrow{c}}{[\overrightarrow{a} \overrightarrow{b} \overrightarrow{c}]}, \overrightarrow{q}=\frac{\overrightarrow{c} \times \overrightarrow{a}}{[\overrightarrow{a} \overrightarrow{b} \overrightarrow{c}]}, \overrightarrow{r}=\frac{\overrightarrow{a} \times \overrightarrow{b}}{[\overrightarrow{a} \overrightarrow{b} \overrightarrow{c}]}$.
We need to calculate $\overrightarrow{a} \cdot \overrightarrow{p}+\overrightarrow{b} \cdot \overrightarrow{q}+\overrightarrow{c} \cdot \overrightarrow{r}$.
Substituting the values,we get:
$\overrightarrow{a} \cdot \overrightarrow{p} = \overrightarrow{a} \cdot \frac{\overrightarrow{b} \times \overrightarrow{c}}{[\overrightarrow{a} \overrightarrow{b} \overrightarrow{c}]} = \frac{[\overrightarrow{a} \overrightarrow{b} \overrightarrow{c}]}{[\overrightarrow{a} \overrightarrow{b} \overrightarrow{c}]} = 1$.
$\overrightarrow{b} \cdot \overrightarrow{q} = \overrightarrow{b} \cdot \frac{\overrightarrow{c} \times \overrightarrow{a}}{[\overrightarrow{a} \overrightarrow{b} \overrightarrow{c}]} = \frac{[\overrightarrow{b} \overrightarrow{c} \overrightarrow{a}]}{[\overrightarrow{a} \overrightarrow{b} \overrightarrow{c}]} = \frac{[\overrightarrow{a} \overrightarrow{b} \overrightarrow{c}]}{[\overrightarrow{a} \overrightarrow{b} \overrightarrow{c}]} = 1$.
$\overrightarrow{c} \cdot \overrightarrow{r} = \overrightarrow{c} \cdot \frac{\overrightarrow{a} \times \overrightarrow{b}}{[\overrightarrow{a} \overrightarrow{b} \overrightarrow{c}]} = \frac{[\overrightarrow{c} \overrightarrow{a} \overrightarrow{b}]}{[\overrightarrow{a} \overrightarrow{b} \overrightarrow{c}]} = \frac{[\overrightarrow{a} \overrightarrow{b} \overrightarrow{c}]}{[\overrightarrow{a} \overrightarrow{b} \overrightarrow{c}]} = 1$.
Therefore,$\overrightarrow{a} \cdot \overrightarrow{p}+\overrightarrow{b} \cdot \overrightarrow{q}+\overrightarrow{c} \cdot \overrightarrow{r} = 1+1+1 = 3$.

(B) The volume of a parallelepiped with coterminous edges represented by vectors $\overrightarrow{OA}$, $\overrightarrow{OB}$, and $\overrightarrow{OC}$ is given by the scalar triple product $|[\overrightarrow{OA} \overrightarrow{OB} \overrightarrow{OC}]|$.
Given the vertices $O(0,0,0)$, $A(2,-2,1)$, $B(5,-4,4)$, and $C(1,-2,4)$, the vectors are:
$\overrightarrow{OA} = 2\hat{i} - 2\hat{j} + \hat{k}$
$\overrightarrow{OB} = 5\hat{i} - 4\hat{j} + 4\hat{k}$
$\overrightarrow{OC} = \hat{i} - 2\hat{j} + 4\hat{k}$
The volume is the absolute value of the determinant:
$V = \left| \begin{vmatrix} 2 & -2 & 1 \\ 5 & -4 & 4 \\ 1 & -2 & 4 \end{vmatrix} \right|$
Expanding along the first row:
$V = |2((-4)(4) - (4)(-2)) - (-2)((5)(4) - (4)(1)) + 1((5)(-2) - (-4)(1))|$
$V = |2(-16 + 8) + 2(20 - 4) + 1(-10 + 4)|$
$V = |2(-8) + 2(16) + 1(-6)|$
$V = |-16 + 32 - 6|$
$V = |10| = 10 \text{ cubic units}$.

(D) The volume of a parallelepiped with coterminal edges $\vec{u}, \vec{v}, \vec{w}$ is given by the scalar triple product $[\vec{u} \vec{v} \vec{w}] = \vec{u} \cdot (\vec{v} \times \vec{w})$.
Given $[\bar{a} \bar{b} \bar{c}] = \bar{a} \cdot (\bar{b} \times \bar{c}) = 4$.
We need to find the volume of the parallelepiped with edges $\vec{u} = \bar{a} + 2 \bar{b}$,$\vec{v} = \bar{b} + 2 \bar{c}$,and $\vec{w} = \bar{c} + 2 \bar{a}$.
Volume $= [(\bar{a} + 2 \bar{b}) (\bar{b} + 2 \bar{c}) (\bar{c} + 2 \bar{a})]$.
Using the property of scalar triple products,$[\bar{a} + 2 \bar{b}, \bar{b} + 2 \bar{c}, \bar{c} + 2 \bar{a}] = [\bar{a} \bar{b} \bar{c}] + 8 [\bar{b} \bar{c} \bar{a}] = [\bar{a} \bar{b} \bar{c}] + 8 [\bar{a} \bar{b} \bar{c}] = 9 [\bar{a} \bar{b} \bar{c}]$.
Substituting the given value: $9 \times 4 = 36$ cubic units.

(D) The scalar triple product of three vectors $a, b, c$ is defined as $a \cdot(b \times c)$.
It follows the cyclic property: $a \cdot(b \times c) = b \cdot(c \times a) = c \cdot(a \times b)$.
Given the expression $w \cdot(u \times v)$,by the cyclic property,this is equal to $u \cdot(v \times w)$ and $v \cdot(w \times u)$.
Also,since the dot product is commutative,$w \cdot(u \times v) = (u \times v) \cdot w$.
However,$v \cdot(u \times w) = -(v \cdot(w \times u)) = -(w \cdot(u \times v))$.
Therefore,$v \cdot(u \times w)$ is not equal to $w \cdot(u \times v)$.

(B) The vectors representing the co-terminus edges are given by:
$AB = (5-3)\hat{i} + (-2-7)\hat{j} + (-3-4)\hat{k} = 2\hat{i} - 9\hat{j} - 7\hat{k}$
$AC = (-4-3)\hat{i} + (5-7)\hat{j} + (6-4)\hat{k} = -7\hat{i} - 2\hat{j} + 2\hat{k}$
$AD = (1-3)\hat{i} + (2-7)\hat{j} + (3-4)\hat{k} = -2\hat{i} - 5\hat{j} - 1\hat{k}$
The volume of the parallelepiped is given by the scalar triple product $|[AB, AC, AD]| = |\vec{AB} \cdot (\vec{AC} \times \vec{AD})|$.
This is equal to the absolute value of the determinant:
$|\begin{vmatrix} 2 & -9 & -7 \\ -7 & -2 & 2 \\ -2 & -5 & -1 \end{vmatrix}|$
$= |2(2 - (-10)) - (-9)(7 - (-4)) + (-7)(35 - 4)|$
$= |2(12) + 9(11) - 7(31)|$
$= |24 + 99 - 217|$
$= |123 - 217| = |-94| = 94$ cubic units.

(C) Given that $\vec{p}, \vec{q}, \vec{r}$ are non-zero,non-coplanar vectors:
The scalar triple product $[\vec{p}+\vec{q}-\vec{r}, \vec{p}-\vec{q}, \vec{q}-\vec{r}]$ can be expressed as the determinant of the coefficients of $\vec{p}, \vec{q}, \vec{r}$ multiplied by the scalar triple product $[\vec{p} \quad \vec{q} \quad \vec{r}]$.
$[\vec{p}+\vec{q}-\vec{r}, \vec{p}-\vec{q}, \vec{q}-\vec{r}] = \begin{vmatrix} 1 & 1 & -1 \\ 1 & -1 & 0 \\ 0 & 1 & -1 \end{vmatrix} [\vec{p} \quad \vec{q} \quad \vec{r}]$
Calculating the determinant:
$= [1((-1)(-1) - (0)(1)) - 1((1)(-1) - (0)(0)) + (-1)((1)(1) - (-1)(0))] [\vec{p} \quad \vec{q} \quad \vec{r}]$
$= [1(1) - 1(-1) - 1(1)] [\vec{p} \quad \vec{q} \quad \vec{r}]$
$= (1 + 1 - 1) [\vec{p} \quad \vec{q} \quad \vec{r}]$
$= [\vec{p} \quad \vec{q} \quad \vec{r}]$

(C) The volume of a parallelepiped with coterminous edges $\vec{u}, \vec{v}, \vec{w}$ is given by the scalar triple product $[\vec{u} \vec{v} \vec{w}] = \vec{u} \cdot (\vec{v} \times \vec{w})$.
Given edges are $\vec{a}+\vec{b}, \vec{b}+\vec{c}, \text{ and } \vec{c}+\vec{a}$.
The volume $V = [(\vec{a}+\vec{b}) (\vec{b}+\vec{c}) (\vec{c}+\vec{a})]$.
Using the property of scalar triple product,$[\vec{a}+\vec{b}, \vec{b}+\vec{c}, \vec{c}+\vec{a}] = (\vec{a}+\vec{b}) \cdot ((\vec{b}+\vec{c}) \times (\vec{c}+\vec{a}))$.
Expanding the cross product: $(\vec{b}+\vec{c}) \times (\vec{c}+\vec{a}) = \vec{b} \times \vec{c} + \vec{b} \times \vec{a} + \vec{c} \times \vec{c} + \vec{c} \times \vec{a}$.
Since $\vec{c} \times \vec{c} = 0$,this simplifies to $\vec{b} \times \vec{c} + \vec{b} \times \vec{a} + \vec{c} \times \vec{a}$.
Now,$V = (\vec{a}+\vec{b}) \cdot (\vec{b} \times \vec{c} + \vec{b} \times \vec{a} + \vec{c} \times \vec{a})$.
$V = \vec{a} \cdot (\vec{b} \times \vec{c}) + \vec{a} \cdot (\vec{b} \times \vec{a}) + \vec{a} \cdot (\vec{c} \times \vec{a}) + \vec{b} \cdot (\vec{b} \times \vec{c}) + \vec{b} \cdot (\vec{b} \times \vec{a}) + \vec{b} \cdot (\vec{c} \times \vec{a})$.
Terms like $\vec{a} \cdot (\vec{b} \times \vec{a})$ are zero because $\vec{a}$ is perpendicular to $\vec{b} \times \vec{a}$.
Only $\vec{a} \cdot (\vec{b} \times \vec{c})$ and $\vec{b} \cdot (\vec{c} \times \vec{a})$ remain.
$V = [\vec{a} \vec{b} \vec{c}] + [\vec{b} \vec{c} \vec{a}] = [\vec{a} \vec{b} \vec{c}] + [\vec{a} \vec{b} \vec{c}] = 2[\vec{a} \vec{b} \vec{c}]$.

(A) The scalar triple product $\vec{a} \cdot (\vec{b} \times \vec{c})$ is given by the determinant of the components of the vectors $\vec{a}$,$\vec{b}$,and $\vec{c}$.
$\vec{a} \cdot (\vec{b} \times \vec{c}) = \begin{vmatrix} 1 & 1 & 1 \\ 2 & \lambda & 1 \\ 1 & -1 & 4 \end{vmatrix} = 10$
Expanding the determinant along the first row:
$1(4\lambda - (-1)) - 1(8 - 1) + 1(-2 - \lambda) = 10$
$1(4\lambda + 1) - 1(7) + 1(-2 - \lambda) = 10$
$4\lambda + 1 - 7 - 2 - \lambda = 10$
$3\lambda - 8 = 10$
$3\lambda = 18$
$\lambda = 6$

(C) We know the properties of unit vectors in cross product:
$\hat{i} \times \hat{j} = \hat{k}$,$\hat{j} \times \hat{k} = \hat{i}$,$\hat{k} \times \hat{i} = \hat{j}$.
Also,$\hat{a} \times \hat{b} = -(\hat{b} \times \hat{a})$ and $\hat{a} \times \hat{a} = 0$.
$1$. $\hat{j} \cdot(\hat{i} \times \hat{k}) = \hat{j} \cdot(-\hat{j}) = -(\hat{j} \cdot \hat{j}) = -1$.
$2$. $\hat{i} \cdot(\hat{j} \times \hat{j}) = \hat{i} \cdot(0) = 0$.
$3$. $\hat{k} \cdot(\hat{j} \times \hat{i}) = \hat{k} \cdot(-\hat{k}) = -(\hat{k} \cdot \hat{k}) = -1$.
$4$. $\hat{i} \cdot(\hat{k} \times \hat{j}) = \hat{i} \cdot(-\hat{i}) = -(\hat{i} \cdot \hat{i}) = -1$.
Summing these values: $(-1) + 0 + (-1) + (-1) = -3$.
Thus,the correct option is $C$.

(NONE) The scalar triple product $\vec{a} \cdot(\vec{b} \times \vec{c})$ is given by the determinant of the matrix formed by the components of vectors $\vec{a}, \vec{b}$,and $\vec{c}$.
$\vec{a} \cdot(\vec{b} \times \vec{c}) = \begin{vmatrix} 2 & -1 & 3 \\ 1 & -2 & 1 \\ 3 & -1 & 2 \end{vmatrix}$
Expanding the determinant along the first row:
$= 2((-2)(2) - (1)(-1)) - (-1)((1)(2) - (1)(3)) + 3((1)(-1) - (-2)(3))$
$= 2(-4 + 1) + 1(2 - 3) + 3(-1 + 6)$
$= 2(-3) + 1(-1) + 3(5)$
$= -6 - 1 + 15$
$= 8$.

(D) The volume of a parallelepiped with coterminous edges $\vec{a}, \vec{b}, \text{ and } \vec{c}$ is given by the scalar triple product $|\vec{a} \cdot (\vec{b} \times \vec{c})|$.
This is equivalent to the absolute value of the determinant of the matrix formed by the components of the vectors:
$V = |\det \begin{bmatrix} 2 & 1 & 1 \\ 3 & -1 & 1 \\ -1 & 1 & -1 \end{bmatrix}|$
Expanding the determinant along the first row:
$V = |2((-1)(-1) - (1)(1)) - 1((3)(-1) - (1)(-1)) + 1((3)(1) - (-1)(-1))|$
$V = |2(1 - 1) - 1(-3 + 1) + 1(3 - 1)|$
$V = |2(0) - 1(-2) + 1(2)|$
$V = |0 + 2 + 2| = |4| = 4$
Thus,the volume is $4$ cubic units.

(C) The scalar triple product is defined as $[\vec{x} \quad \vec{y} \quad \vec{z}] = \vec{x} \cdot (\vec{y} \times \vec{z})$.
Given expression: $[\vec{a}-\vec{b} \quad \vec{b}-\vec{c} \quad \vec{c}-\vec{a}] = (\vec{a}-\vec{b}) \cdot ((\vec{b}-\vec{c}) \times (\vec{c}-\vec{a}))$.
First,calculate the cross product: $(\vec{b}-\vec{c}) \times (\vec{c}-\vec{a}) = \vec{b} \times \vec{c} - \vec{b} \times \vec{a} - \vec{c} \times \vec{c} + \vec{c} \times \vec{a}$.
Since $\vec{c} \times \vec{c} = 0$,this simplifies to $\vec{b} \times \vec{c} + \vec{a} \times \vec{b} + \vec{c} \times \vec{a}$.
Now,take the dot product with $(\vec{a}-\vec{b})$:
$(\vec{a}-\vec{b}) \cdot (\vec{b} \times \vec{c} + \vec{a} \times \vec{b} + \vec{c} \times \vec{a})$
$= \vec{a} \cdot (\vec{b} \times \vec{c}) + \vec{a} \cdot (\vec{a} \times \vec{b}) + \vec{a} \cdot (\vec{c} \times \vec{a}) - \vec{b} \cdot (\vec{b} \times \vec{c}) - \vec{b} \cdot (\vec{a} \times \vec{b}) - \vec{b} \cdot (\vec{c} \times \vec{a})$.
Using the property that the scalar triple product is zero if any two vectors are identical:
$= [\vec{a} \vec{b} \vec{c}] + 0 + 0 - 0 - 0 - [\vec{b} \vec{c} \vec{a}]$.
Since $[\vec{a} \vec{b} \vec{c}] = [\vec{b} \vec{c} \vec{a}]$,the expression becomes $[\vec{a} \vec{b} \vec{c}] - [\vec{a} \vec{b} \vec{c}] = 0$.

(B) For three vectors to be coplanar,their scalar triple product must be zero.
Given the vectors $\vec{u} = a\hat{i}+\hat{j}+\hat{k}$,$\vec{v} = \hat{i}+b\hat{j}+\hat{k}$,and $\vec{w} = \hat{i}+\hat{j}+c\hat{k}$.
The condition for coplanarity is:
$\begin{vmatrix} a & 1 & 1 \\ 1 & b & 1 \\ 1 & 1 & c \end{vmatrix} = 0$
Expanding the determinant along the first row:
$a(bc - 1) - 1(c - 1) + 1(1 - b) = 0$
$abc - a - c + 1 + 1 - b = 0$
$abc - a - b - c + 2 = 0$
$abc - (a + b + c) = -2$

(B) The volume of a parallelopiped with coterminous edges $\vec{a}, \vec{b}, \vec{c}$ is given by the scalar triple product $|\vec{a} \cdot (\vec{b} \times \vec{c})|$, which is equivalent to the determinant of the matrix formed by the components of the vectors.
Given vectors are $\vec{a} = 0\hat{i} + 1\hat{j} + 1\hat{k}$, $\vec{b} = 1\hat{i} + 0\hat{j} + 1\hat{k}$, and $\vec{c} = 1\hat{i} + 1\hat{j} + 0\hat{k}$.
The volume $V = \left|\begin{array}{lll}0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0\end{array}\right|$.
Expanding the determinant along the first row:
$V = |0(0 - 1) - 1(0 - 1) + 1(1 - 0)|$.
$V = |0 + 1 + 1| = |2| = 2 \text{ cu units}$.

(B) The volume of a parallelepiped formed by vectors $\vec{a}, \vec{b}, \vec{c}$ is given by the scalar triple product $[\vec{a} \vec{b} \vec{c}] = 4$.
We need to find the value of $[\vec{a} \times \vec{b} \quad \vec{b} \times \vec{c} \quad \vec{c} \times \vec{a}]$.
Using the property of the scalar triple product of cross products: $[\vec{a} \times \vec{b} \quad \vec{b} \times \vec{c} \quad \vec{c} \times \vec{a}] = [\vec{a} \vec{b} \vec{c}]^2$.
Substituting the given value: $[\vec{a} \vec{b} \vec{c}]^2 = (4)^2 = 16$.
Thus,the value is $16$.

(C) Given that $a, b, c$ are non-coplanar,so the scalar triple product $[a b c] \neq 0$.
We know that $[a b c] = [b c a] = [c a b]$.
Let $V = [a b c]$.
The expression is $E = a \cdot \left\{ \frac{b \times c}{3 b \cdot (c \times a)} \right\} - b \cdot \left\{ \frac{c \times a}{2 c \cdot (a \times b)} \right\}$.
Using the definition of scalar triple product,$b \cdot (c \times a) = [b c a] = [a b c] = V$ and $c \cdot (a \times b) = [c a b] = [a b c] = V$.
Substituting these into the expression:
$E = \frac{a \cdot (b \times c)}{3 [a b c]} - \frac{b \cdot (c \times a)}{2 [a b c]}$
$E = \frac{[a b c]}{3 [a b c]} - \frac{[b c a]}{2 [a b c]}$
Since $[a b c] = [b c a] = V$,we have:
$E = \frac{V}{3V} - \frac{V}{2V} = \frac{1}{3} - \frac{1}{2} = \frac{2 - 3}{6} = -\frac{1}{6}$.

(D) Let the vectors be $\vec{a} = 2i + 3j + 0k$,$\vec{b} = i + j + k$,and $\vec{c} = \lambda i + 4j + 2k$.
The volume of a parallelepiped with coterminous edges $\vec{a}, \vec{b}, \vec{c}$ is given by the absolute value of the scalar triple product $|[\vec{a} \vec{b} \vec{c}]| = |(\vec{a} \times \vec{b}) \cdot \vec{c}|$.
First,calculate the cross product $\vec{a} \times \vec{b}$:
$\vec{a} \times \vec{b} = \begin{vmatrix} i & j & k \\ 2 & 3 & 0 \\ 1 & 1 & 1 \end{vmatrix} = i(3-0) - j(2-0) + k(2-3) = 3i - 2j - k$.
Now,calculate the scalar triple product:
$(\vec{a} \times \vec{b}) \cdot \vec{c} = (3i - 2j - k) \cdot (\lambda i + 4j + 2k) = 3\lambda - 8 - 2 = 3\lambda - 10$.
Given the volume is $2$,we have $|3\lambda - 10| = 2$.
Case $1$: $3\lambda - 10 = 2 \Rightarrow 3\lambda = 12 \Rightarrow \lambda = 4$.
Case $2$: $3\lambda - 10 = -2 \Rightarrow 3\lambda = 8 \Rightarrow \lambda = 8/3$.
Since $4$ is the provided option,the value is $4$.

(B) Given that $\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c}$ are non-zero coplanar vectors,their scalar triple product is zero,i.e.,$[\overrightarrow{a} \overrightarrow{b} \overrightarrow{c}] = 0$.
We need to evaluate the scalar triple product $[2 \overrightarrow{a}-\overrightarrow{b} \quad 3 \overrightarrow{b}-\overrightarrow{c} \quad 4 \overrightarrow{c}-\overrightarrow{a}]$.
Using the definition of the scalar triple product:
$[2 \overrightarrow{a}-\overrightarrow{b} \quad 3 \overrightarrow{b}-\overrightarrow{c} \quad 4 \overrightarrow{c}-\overrightarrow{a}] = (2 \overrightarrow{a}-\overrightarrow{b}) \cdot [(3 \overrightarrow{b}-\overrightarrow{c}) \times (4 \overrightarrow{c}-\overrightarrow{a})]$
$= (2 \overrightarrow{a}-\overrightarrow{b}) \cdot [12(\overrightarrow{b} \times \overrightarrow{c}) - 3(\overrightarrow{b} \times \overrightarrow{a}) - 4(\overrightarrow{c} \times \overrightarrow{c}) + (\overrightarrow{c} \times \overrightarrow{a})]$
Since $\overrightarrow{c} \times \overrightarrow{c} = 0$,the expression simplifies to:
$= (2 \overrightarrow{a}-\overrightarrow{b}) \cdot [12(\overrightarrow{b} \times \overrightarrow{c}) - 3(\overrightarrow{b} \times \overrightarrow{a}) + (\overrightarrow{c} \times \overrightarrow{a})]$
$= 24[\overrightarrow{a} \overrightarrow{b} \overrightarrow{c}] - 6[\overrightarrow{a} \overrightarrow{b} \overrightarrow{a}] + 2[\overrightarrow{a} \overrightarrow{c} \overrightarrow{a}] - 12[\overrightarrow{b} \overrightarrow{b} \overrightarrow{c}] + 3[\overrightarrow{b} \overrightarrow{b} \overrightarrow{a}] - [\overrightarrow{b} \overrightarrow{c} \overrightarrow{a}]$
Since any scalar triple product with two identical vectors is zero,$[\overrightarrow{a} \overrightarrow{b} \overrightarrow{a}] = 0, [\overrightarrow{a} \overrightarrow{c} \overrightarrow{a}] = 0, [\overrightarrow{b} \overrightarrow{b} \overrightarrow{c}] = 0, [\overrightarrow{b} \overrightarrow{b} \overrightarrow{a}] = 0$.
Thus,the expression becomes:
$= 24[\overrightarrow{a} \overrightarrow{b} \overrightarrow{c}] - [\overrightarrow{b} \overrightarrow{c} \overrightarrow{a}]$
Since $[\overrightarrow{a} \overrightarrow{b} \overrightarrow{c}] = [\overrightarrow{b} \overrightarrow{c} \overrightarrow{a}]$,we have:
$= 24[\overrightarrow{a} \overrightarrow{b} \overrightarrow{c}] - [\overrightarrow{a} \overrightarrow{b} \overrightarrow{c}] = 23[\overrightarrow{a} \overrightarrow{b} \overrightarrow{c}]$
Since $\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c}$ are coplanar,$[\overrightarrow{a} \overrightarrow{b} \overrightarrow{c}] = 0$.
Therefore,$23 \times 0 = 0$.

(A) Given,the volume of the parallelepiped with coterminous edges $\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c}$ is given by the scalar triple product $[\overrightarrow{a} \overrightarrow{b} \overrightarrow{c}] = 40 \text{ cubic units}$.
The volume of the parallelepiped with coterminous edges $\overrightarrow{b}+\overrightarrow{c}, \overrightarrow{c}+\overrightarrow{a}, \overrightarrow{a}+\overrightarrow{b}$ is given by the scalar triple product $[\overrightarrow{b}+\overrightarrow{c}, \overrightarrow{c}+\overrightarrow{a}, \overrightarrow{a}+\overrightarrow{b}]$.
Using the property of the scalar triple product:
$[\overrightarrow{b}+\overrightarrow{c}, \overrightarrow{c}+\overrightarrow{a}, \overrightarrow{a}+\overrightarrow{b}] = (\overrightarrow{b}+\overrightarrow{c}) \cdot ((\overrightarrow{c}+\overrightarrow{a}) \times (\overrightarrow{a}+\overrightarrow{b}))$
$= (\overrightarrow{b}+\overrightarrow{c}) \cdot (\overrightarrow{c} \times \overrightarrow{a} + \overrightarrow{c} \times \overrightarrow{b} + \overrightarrow{a} \times \overrightarrow{a} + \overrightarrow{a} \times \overrightarrow{b})$
Since $\overrightarrow{a} \times \overrightarrow{a} = 0$,this simplifies to:
$= (\overrightarrow{b}+\overrightarrow{c}) \cdot (\overrightarrow{c} \times \overrightarrow{a} + \overrightarrow{c} \times \overrightarrow{b} + \overrightarrow{a} \times \overrightarrow{b})$
$= \overrightarrow{b} \cdot (\overrightarrow{c} \times \overrightarrow{a}) + \overrightarrow{b} \cdot (\overrightarrow{c} \times \overrightarrow{b}) + \overrightarrow{b} \cdot (\overrightarrow{a} \times \overrightarrow{b}) + \overrightarrow{c} \cdot (\overrightarrow{c} \times \overrightarrow{a}) + \overrightarrow{c} \cdot (\overrightarrow{c} \times \overrightarrow{b}) + \overrightarrow{c} \cdot (\overrightarrow{a} \times \overrightarrow{b})$
$= [\overrightarrow{b} \overrightarrow{c} \overrightarrow{a}] + 0 + 0 + 0 + 0 + [\overrightarrow{c} \overrightarrow{a} \overrightarrow{b}]$
$= [\overrightarrow{a} \overrightarrow{b} \overrightarrow{c}] + [\overrightarrow{a} \overrightarrow{b} \overrightarrow{c}] = 2[\overrightarrow{a} \overrightarrow{b} \overrightarrow{c}]$
$= 2 \times 40 = 80 \text{ cubic units}$.

(D) Given $p=\frac{b \times c}{[a b c]}, q=\frac{c \times a}{[a b c]}, r=\frac{a \times b}{[a b c]}$.
We need to evaluate the expression $E = (a+b) \cdot p + (b+c) \cdot q + (c+a) \cdot r$.
Substituting the values of $p, q, r$:
$E = (a+b) \cdot \frac{b \times c}{[a b c]} + (b+c) \cdot \frac{c \times a}{[a b c]} + (c+a) \cdot \frac{a \times b}{[a b c]}$
$E = \frac{1}{[a b c]} [a \cdot (b \times c) + b \cdot (b \times c) + b \cdot (c \times a) + c \cdot (c \times a) + c \cdot (a \times b) + a \cdot (a \times b)]$
Using the property of scalar triple product $[x y z] = x \cdot (y \times z)$:
$E = \frac{1}{[a b c]} [[a b c] + 0 + [b c a] + 0 + [c a b] + 0]$
Since $[a b c] = [b c a] = [c a b]$:
$E = \frac{[a b c] + [a b c] + [a b c]}{[a b c]} = \frac{3[a b c]}{[a b c]} = 3$.

(B) The scalar triple product is defined as $[\vec{x}, \vec{y}, \vec{z}] = \vec{x} \cdot (\vec{y} \times \vec{z})$.
Let $\vec{x} = \vec{a}+2 \vec{b}-\vec{c}$,$\vec{y} = \vec{a}-\vec{b}$,and $\vec{z} = \vec{a}-\vec{b}-\vec{c}$.
We calculate the cross product $\vec{y} \times \vec{z} = (\vec{a}-\vec{b}) \times (\vec{a}-\vec{b}-\vec{c})$.
$= \vec{a} \times \vec{a} - \vec{a} \times \vec{b} - \vec{a} \times \vec{c} - \vec{b} \times \vec{a} + \vec{b} \times \vec{b} + \vec{b} \times \vec{c}$.
Since $\vec{a} \times \vec{a} = 0$ and $\vec{b} \times \vec{b} = 0$,and $\vec{b} \times \vec{a} = -(\vec{a} \times \vec{b})$,we get:
$= 0 - (\vec{a} \times \vec{b}) - (\vec{a} \times \vec{c}) + (\vec{a} \times \vec{b}) + 0 + (\vec{b} \times \vec{c}) = \vec{b} \times \vec{c} - \vec{a} \times \vec{c}$.
Now,calculate the dot product $(\vec{a}+2 \vec{b}-\vec{c}) \cdot (\vec{b} \times \vec{c} - \vec{a} \times \vec{c})$.
$= \vec{a} \cdot (\vec{b} \times \vec{c}) - \vec{a} \cdot (\vec{a} \times \vec{c}) + 2\vec{b} \cdot (\vec{b} \times \vec{c}) - 2\vec{b} \cdot (\vec{a} \times \vec{c}) - \vec{c} \cdot (\vec{b} \times \vec{c}) + \vec{c} \cdot (\vec{a} \times \vec{c})$.
$= [\vec{a}, \vec{b}, \vec{c}] - 0 + 0 - 2[\vec{b}, \vec{a}, \vec{c}] - 0 + 0$.
Since $[\vec{b}, \vec{a}, \vec{c}] = -[\vec{a}, \vec{b}, \vec{c}]$,we have:
$= [\vec{a}, \vec{b}, \vec{c}] - 2(-[\vec{a}, \vec{b}, \vec{c}]) = [\vec{a}, \vec{b}, \vec{c}] + 2[\vec{a}, \vec{b}, \vec{c}] = 3[\vec{a}, \vec{b}, \vec{c}]$.

(NONE) Three vectors $\vec{a}, \vec{b}, \vec{c}$ are coplanar if their scalar triple product is zero,i.e.,$[\vec{a} \vec{b} \vec{c}] = 0$.
The given vectors are $\vec{a} = 2 \hat{i} + 3 \hat{j} + 4 \hat{k}$,$\vec{b} = 2 \hat{i} + \hat{j} - \hat{k}$,and $\vec{c} = \lambda \hat{i} - \hat{j} + 2 \hat{k}$.
The condition for coplanarity is the determinant of the components being zero:
$\begin{vmatrix} 2 & 3 & 4 \\ 2 & 1 & -1 \\ \lambda & -1 & 2 \end{vmatrix} = 0$
Expanding along the first row:
$2(1(2) - (-1)(-1)) - 3(2(2) - (-1)(\lambda)) + 4(2(-1) - 1(\lambda)) = 0$
$2(2 - 1) - 3(4 + \lambda) + 4(-2 - \lambda) = 0$
$2(1) - 12 - 3\lambda - 8 - 4\lambda = 0$
$2 - 12 - 8 - 7\lambda = 0$
$-18 - 7\lambda = 0$
$-7\lambda = 18$
$\lambda = -\frac{18}{7}$

(C) Three vectors $\vec{a}$,$\vec{b}$,and $\vec{c}$ are coplanar if and only if their scalar triple product is zero,i.e.,$[\vec{a} \vec{b} \vec{c}] = 0$.
This is equivalent to the determinant of the matrix formed by their components being zero:
$\begin{vmatrix} 2 & -1 & 3 \\ 1 & 4 & 1 \\ 4 & p & 1 \end{vmatrix} = 0$
Expanding the determinant along the first row:
$2(4(1) - 1(p)) - (-1)(1(1) - 1(4)) + 3(1(p) - 4(4)) = 0$
$2(4 - p) + 1(1 - 4) + 3(p - 16) = 0$
$8 - 2p - 3 + 3p - 48 = 0$
$p - 43 = 0$
$p = 43$
Thus,the correct option is $C$.

(D) Three vectors $\vec{a}, \vec{b}, \vec{c}$ are coplanar if their scalar triple product is zero,i.e.,$[\vec{a} \vec{b} \vec{c}] = 0$.
Given vectors are $\vec{a} = 2 \bar{i} + 4 \bar{j} - 3 \bar{k}$,$\vec{b} = -\bar{i} + 2 \bar{j} + 3 \bar{k}$,and $\vec{c} = p \bar{i} - 2 \bar{j} + \bar{k}$.
The condition $[\vec{a} \vec{b} \vec{c}] = 0$ implies the determinant of the components is zero:
$\begin{vmatrix} 2 & 4 & -3 \\ -1 & 2 & 3 \\ p & -2 & 1 \end{vmatrix} = 0$
Expanding along the first row:
$2(2(1) - 3(-2)) - 4(-1(1) - 3(p)) - 3(-1(-2) - 2(p)) = 0$
$2(2 + 6) - 4(-1 - 3p) - 3(2 - 2p) = 0$
$2(8) + 4 + 12p - 6 + 6p = 0$
$16 + 4 - 6 + 18p = 0$
$14 + 18p = 0 \implies 18p = -14 \implies p = -\frac{7}{9}$.
Now,substitute $p = -\frac{7}{9}$ into the vector $9p \bar{i} - 4 \bar{j} + 4 \bar{k}$:
$9(-\frac{7}{9}) \bar{i} - 4 \bar{j} + 4 \bar{k} = -7 \bar{i} - 4 \bar{j} + 4 \bar{k}$.
Let $\vec{v} = -7 \bar{i} - 4 \bar{j} + 4 \bar{k}$. The magnitude is $|\vec{v}| = \sqrt{(-7)^2 + (-4)^2 + 4^2} = \sqrt{49 + 16 + 16} = \sqrt{81} = 9$.
The unit vector is $\frac{\vec{v}}{|\vec{v}|} = \frac{1}{9}(-7 \bar{i} - 4 \bar{j} + 4 \bar{k})$.

(C) Three vectors $\vec{a}, \vec{b},$ and $\vec{c}$ are coplanar if and only if their scalar triple product is zero,i.e.,$[\vec{a} \vec{b} \vec{c}] = 0$.
This is equivalent to the determinant of the matrix formed by their components being zero:
$\left|\begin{array}{ccc} 2 & -1 & 1 \\ 1 & 2 & -3 \\ 3 & p & 5 \end{array}\right| = 0$
Expanding the determinant along the first row:
$2(2 \times 5 - (-3) \times p) - (-1)(1 \times 5 - (-3) \times 3) + 1(1 \times p - 2 \times 3) = 0$
$2(10 + 3p) + 1(5 + 9) + 1(p - 6) = 0$
$20 + 6p + 14 + p - 6 = 0$
$7p + 28 = 0$
$7p = -28$
$p = -4$

(C) Let the position vectors of the points be $\overrightarrow{OA} = -\hat{i} + 4\hat{j} - 4\hat{k}$,$\overrightarrow{OB} = 3\hat{i} + 2\hat{j} - 5\hat{k}$,$\overrightarrow{OC} = -3\hat{i} + 8\hat{j} - 5\hat{k}$,and $\overrightarrow{OD} = -3\hat{i} + 2\hat{j} + \lambda\hat{k}$.
For the points to be coplanar,the scalar triple product of the vectors $\overrightarrow{AB}$,$\overrightarrow{AC}$,and $\overrightarrow{AD}$ must be zero.
First,calculate the vectors:
$\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = (3 - (-1))\hat{i} + (2 - 4)\hat{j} + (-5 - (-4))\hat{k} = 4\hat{i} - 2\hat{j} - \hat{k}$
$\overrightarrow{AC} = \overrightarrow{OC} - \overrightarrow{OA} = (-3 - (-1))\hat{i} + (8 - 4)\hat{j} + (-5 - (-4))\hat{k} = -2\hat{i} + 4\hat{j} - \hat{k}$
$\overrightarrow{AD} = \overrightarrow{OD} - \overrightarrow{OA} = (-3 - (-1))\hat{i} + (2 - 4)\hat{j} + (\lambda - (-4))\hat{k} = -2\hat{i} - 2\hat{j} + (\lambda + 4)\hat{k}$
Now,set the determinant of these vectors to zero:
$\begin{vmatrix} 4 & -2 & -1 \\ -2 & 4 & -1 \\ -2 & -2 & \lambda + 4 \end{vmatrix} = 0$
Expanding the determinant:
$4(4(\lambda + 4) - 2) - (-2)(-2(\lambda + 4) - 2) - 1(4 - (-8)) = 0$
$4(4\lambda + 16 - 2) + 2(-2\lambda - 8 - 2) - 1(12) = 0$
$4(4\lambda + 14) + 2(-2\lambda - 10) - 12 = 0$
$16\lambda + 56 - 4\lambda - 20 - 12 = 0$
$12\lambda + 24 = 0$
$12\lambda = -24$
$\lambda = -2$

(A) Since the vectors $\vec{a}, \vec{b}, \vec{c}$ are linearly dependent,their scalar triple product must be zero:
$\begin{vmatrix} \alpha & \beta & 3 \\ 0 & 1 & 2 \\ 3 & 2 & 1 \end{vmatrix} = 0$
Expanding along the first row:
$\alpha(1 - 4) - \beta(0 - 6) + 3(0 - 3) = 0$
$-3\alpha + 6\beta - 9 = 0$
Dividing by $-3$:
$\alpha - 2\beta + 3 = 0 \Rightarrow \alpha = 2\beta - 3$
Given that the magnitude $|\vec{a}| = \sqrt{14}$,we have:
$\alpha^2 + \beta^2 + 3^2 = 14$
$\alpha^2 + \beta^2 = 5$
Substituting $\alpha = 2\beta - 3$:
$(2\beta - 3)^2 + \beta^2 = 5$
$4\beta^2 - 12\beta + 9 + \beta^2 = 5$
$5\beta^2 - 12\beta + 4 = 0$
Factoring the quadratic equation:
$(5\beta - 2)(\beta - 2) = 0$
Since $\beta$ is an integer,we take $\beta = 2$.
Then $\alpha = 2(2) - 3 = 1$.
Therefore,$\alpha + \beta = 1 + 2 = 3$.

(A) Given $\vec{a}=\hat{i}-2 \hat{j}$,$\vec{b}=2 \hat{j}+3 \hat{k}$,$\vec{c}=p \hat{i}+q \hat{j}$ and $\vec{d}=p \hat{j}-q \hat{k}$.
First,calculate the cross product $\vec{a} \times \vec{b}$:
$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 0 \\ 0 & 2 & 3 \end{vmatrix} = \hat{i}(-6-0) - \hat{j}(3-0) + \hat{k}(2-0) = -6 \hat{i} - 3 \hat{j} + 2 \hat{k}$.
Now,use the condition $(\vec{a} \times \vec{b}) \cdot \vec{c} = 3$:
$(-6 \hat{i} - 3 \hat{j} + 2 \hat{k}) \cdot (p \hat{i} + q \hat{j}) = 3 \implies -6p - 3q = 3 \implies -2p - q = 1$ (Equation $1$).
Next,use the condition $(\vec{a} \times \vec{b}) \cdot \vec{d} = 3$:
$(-6 \hat{i} - 3 \hat{j} + 2 \hat{k}) \cdot (p \hat{j} - q \hat{k}) = 3 \implies -3p - 2q = 3$ (Equation $2$).
Multiply Equation $1$ by $2$: $-4p - 2q = 2$ (Equation $3$).
Subtract Equation $2$ from Equation $3$: $(-4p - 2q) - (-3p - 2q) = 2 - 3 \implies -p = -1 \implies p = 1$.
Substitute $p=1$ into Equation $1$: $-2(1) - q = 1 \implies -2 - q = 1 \implies q = -3$.
Finally,calculate $3p + q = 3(1) + (-3) = 3 - 3 = 0$.

(B) Three vectors $\vec{a}, \vec{b}, \vec{c}$ are coplanar if their scalar triple product is zero,i.e.,$[\vec{a} \vec{b} \vec{c}] = 0$.
The given vectors are $\vec{a} = 35 \hat{i}+14 \hat{j}-77 \hat{k}$,$\vec{b} = 2 \hat{i}+7 \hat{j}+5 \hat{k}$,and $\vec{c} = 5 \hat{i}+2 \hat{j}+\lambda \hat{k}$.
The condition for coplanarity is given by the determinant:
$\left|\begin{array}{ccc} 35 & 14 & -77 \\ 2 & 7 & 5 \\ 5 & 2 & \lambda \end{array}\right| = 0$
Expanding the determinant along the first row:
$35(7\lambda - 10) - 14(2\lambda - 25) - 77(4 - 35) = 0$
$245\lambda - 350 - 28\lambda + 350 - 77(-31) = 0$
$217\lambda + 2387 = 0$
Dividing by $217$:
$\lambda + 11 = 0$
$\lambda = -11$