(A) Root(s) of the expression $2{\sin ^2}\theta + {\sin ^2}2\theta = 2$

1. The given equation is $2{\sin ^2}\theta + {\sin ^2}2\theta = 2$.
2. Use the double angle identity $\sin(2\theta) = 2\sin\theta\cos\theta$. The equation becomes: $2{\sin ^2}\theta + (2\sin\theta\cos\theta)^2 = 2$ $2{\sin ^2}\theta + 4{\sin ^2}\theta{\cos ^2}\theta = 2$
3. Divide the equation by 2: ${\sin ^2}\theta + 2{\sin ^2}\theta{\cos ^2}\theta = 1$
4. Use the identity ${\cos ^2}\theta = 1 - {\sin ^2}\theta$: ${\sin ^2}\theta + 2{\sin ^2}\theta(1 - {\sin ^2}\theta) = 1$ ${\sin ^2}\theta + 2{\sin ^2}\theta - 2{\sin ^4}\theta = 1$ $3{\sin ^2}\theta - 2{\sin ^4}\theta = 1$
5. Let $x = {\sin ^2}\theta$. The equation becomes a quadratic in $x$: $2x^2 - 3x + 1 = 0$
6. Factor the quadratic equation: $(2x - 1)(x - 1) = 0$
7. The solutions for $x$ are $x = 1/2$ or $x = 1$.
8. Case 1: $x = {\sin ^2}\theta = 1/2$. This gives $\sin\theta = \pm 1/\sqrt{2}$. The principal values for $\theta$ are $\pm \pi/4$. A general solution is $\theta = n\pi \pm \pi/4$. This matches option (Q) ${\pi / 4}$.
9. Case 2: $x = {\sin ^2}\theta = 1$. This gives $\sin\theta = \pm 1$. The principal values for $\theta$ are $\pm \pi/2$. A general solution is $\theta = n\pi + \pi/2$. This matches option (S) ${\pi / 2}$.
10. Therefore, (A) maps to (Q) and (S).

(B) Points of discontinuity of the function $f(x) = \left[ {{{6x} \over \pi }} \right]\cos \left[ {{{3x} \over \pi }} \right]$

1. The greatest integer function $[y]$ is discontinuous at every integer value of $y$. The cosine function is continuous everywhere.
2. The function $f(x)$ can be discontinuous when the arguments of the greatest integer functions are integers. That is, when $\frac{6x}{\pi} = k$ or $\frac{3x}{\pi} = m$ for integers $k, m$.
3. The set of points where $\frac{3x}{\pi} = m$ is a subset of the points where $\frac{6x}{\pi} = k$ (since if $x=m\pi/3$, then $6x/\pi = 2m$, which is an integer). So we only need to check the points $x = \frac{k\pi}{6}$ for $k \in \mathbb{Z}$.
4. Let's check the given values:
   • (P) $x = {\pi / 6}$: Here $k=1$. This is a potential point of discontinuity.
   • (Q) $x = {\pi / 4}$: Here $\frac{6x}{\pi} = 1.5$ and $\frac{3x}{\pi} = 0.75$. Neither is an integer, so $f(x)$ is continuous at $x = \pi/4$.
   • (R) $x = {\pi / 3}$: This is $x = 2\pi/6$, so $k=2$. This is a potential point of discontinuity.
   • (S) $x = {\pi / 2}$: This is $x = 3\pi/6$, so $k=3$. This is a potential point of discontinuity.
   • (T) $x = \pi$: This is $x = 6\pi/6$, so $k=6$. This is a potential point of discontinuity.
5. Let's check for continuity at $x_0 = k\pi/6$. We compare the left-hand limit (LHL) and right-hand limit (RHL). LHL = $\lim_{h \to 0^+} f(x_0 - h) = (k-1)\cos([\frac{k}{2}-\epsilon])$. RHL = $\lim_{h \to 0^+} f(x_0 + h) = k\cos([\frac{k}{2}])$. For continuity, LHL = RHL.
   • For $x=\pi/6$ ($k=1$): LHL = $0 \cdot \cos(0) = 0$. RHL = $1 \cdot \cos(0) = 1$. LHL $\neq$ RHL. Discontinuous.
   • For $x=\pi/3$ ($k=2$): LHL = $(2-1)\cos(1-1) = 1$. RHL = $2\cos(1)$. LHL $\neq$ RHL. Discontinuous.
   • For $x=\pi/2$ ($k=3$): LHL = $(3-1)\cos(1) = 2\cos(1)$. RHL = $3\cos(1)$. LHL $\neq$ RHL. Discontinuous.
   • For $x=\pi$ ($k=6$): LHL = $(6-1)\cos(3-1) = 5\cos(2)$. RHL = $6\cos(3)$. LHL $\neq$ RHL. Discontinuous.
6. So, the points of discontinuity from the given options are $\pi/6, \pi/3, \pi/2, \pi$.
7. Therefore, (B) maps to (P), (R), (S), (T).

(C) Volume of the parallelopiped

1. The question has a likely typo and should list three vectors for the edges. Assuming the vectors are $\vec{a} = \widehat i + \widehat j$, $\vec{b} = \widehat i + 2\widehat j$, and $\vec{c} = \widehat i + \widehat j + \pi \widehat k$.
2. The volume $V$ of the parallelopiped is the absolute value of the scalar triple product $[\vec{a} \vec{b} \vec{c}]$, which can be calculated as the determinant of the matrix formed by the vector components. $\vec{a} = (1, 1, 0), \vec{b} = (1, 2, 0), \vec{c} = (1, 1, \pi)$ $V = \left| \det \begin{pmatrix} 1 & 1 & 0 \\ 1 & 2 & 0 \\ 1 & 1 & \pi \end{pmatrix} \right|$
3. Expanding the determinant along the third column: $V = |\pi (1 \cdot 2 - 1 \cdot 1)| = |\pi (2-1)| = |\pi| = \pi$
4. The volume is $\pi$, which corresponds to option (T).
5. Therefore, (C) maps to (T).

(D) Angle between vectors $\overrightarrow a$ and $\overrightarrow b$

1. We are given that $\vec{a}$, $\vec{b}$, and $\vec{c}$ are unit vectors, so $|\vec{a}| = |\vec{b}| = |\vec{c}| = 1$.
2. The given relation is $\overrightarrow a + \overrightarrow b + \sqrt 3 \overrightarrow c = \overrightarrow 0$.
3. To find the angle $\theta$ between $\vec{a}$ and $\vec{b}$, we isolate these vectors: $\overrightarrow a + \overrightarrow b = -\sqrt 3 \overrightarrow c$
4. Take the dot product of each side with itself (i.e., square the magnitude): $(\overrightarrow a + \overrightarrow b) \cdot (\overrightarrow a + \overrightarrow b) = (-\sqrt 3 \overrightarrow c) \cdot (-\sqrt 3 \overrightarrow c)$ $|\overrightarrow a + \overrightarrow b|^2 = 3|\overrightarrow c|^2$
5. Expand the left side: $|\vec{a}|^2 + |\vec{b}|^2 + 2\vec{a} \cdot \vec{b} = 3|\vec{c}|^2$
6. Substitute the magnitudes and the definition of the dot product $\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos\theta$: $1^2 + 1^2 + 2(1)(1)\cos\theta = 3(1^2)$ $2 + 2\cos\theta = 3$ $2\cos\theta = 1$ $\cos\theta = {1 \over 2}$
7. The angle $\theta$ in the range $[0, \pi]$ is $\theta = \pi/3$.
8. This corresponds to option (R).
9. Therefore, (D) maps to (R).

Conclusion

The final matching is:

• (A) $\to$ (Q), (S)
• (B) $\to$ (P), (R), (S), (T)
• (C) $\to$ (T)
• (D) $\to$ (R) This corresponds to option C in the list of choices.