Condition for Right Angle

Given,

$$|\vec a+\vec b|=\sqrt{21}$$

Therefore,

$$|\vec a+\vec b|^2=21$$

$$|\vec a|^2+|\vec b|^2+2\vec a\cdot\vec b=21 \qquad (1)$$

Also,

$$|\vec a-\vec b|=3$$

Therefore,

$$|\vec a-\vec b|^2=9$$

$$|\vec a|^2+|\vec b|^2-2\vec a\cdot\vec b=9 \qquad (2)$$

Adding (1) and (2),

$$|\vec a|^2+|\vec b|^2=15 \qquad (3)$$

Subtracting (2) from (1),

$$4\vec a\cdot\vec b=12$$

$$\vec a\cdot\vec b=3 \qquad (4)$$

Since

$$\vec a\perp(\vec a-\vec b)$$

we have

$$\vec a\cdot(\vec a-\vec b)=0$$

$$|\vec a|^2-\vec a\cdot\vec b=0$$

Using (4),

$$|\vec a|^2=3$$

Hence from (3),

$$|\vec b|^2=12$$

Now,

$$\text{Area of } \triangle OPR=\frac12|\vec a\times(\vec a+\vec b)|$$

$$=\frac12|\vec a\times\vec b|$$

Also,

$$|\vec a\times\vec b|^2=|\vec a|^2|\vec b|^2-(\vec a\cdot\vec b)^2$$

$$=(3)(12)-3^2$$

$$=36-9$$

$$=27$$

Therefore,

$$|\vec a\times\vec b|=3\sqrt3$$

Hence,

$$\text{Area of } \triangle OPR=\frac12(3\sqrt3)$$

$$=\boxed{\frac{3\sqrt3}{2}}$$

$$M = \begin{bmatrix} \alpha & \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{3}} & \beta & \frac{1}{\sqrt{3}} \\ \gamma & \delta & \mu \end{bmatrix}$$

$$\vec{u} = \alpha\hat{i} + \frac{1}{\sqrt{3}}\hat{j} + \gamma\hat{k}$$

$$\vec{v} = \frac{1}{\sqrt{2}}\hat{i} + \beta\hat{j} + \delta\hat{k}$$

$$\vec{w} = -\frac{1}{\sqrt{2}}\hat{i} + \frac{1}{\sqrt{3}}\hat{j} + \mu\hat{k}$$

$$\alpha^2 + \frac{1}{3} + \gamma^2 = 1 \quad \therefore \alpha^2 + \gamma^2 = \frac{2}{3} \quad \dots(1)$$

$$\frac{1}{2} + \beta^2 + \delta^2 = 1 \quad \beta^2 + \delta^2 = \frac{1}{2} \quad \dots(2)$$

$$\frac{1}{2} + \frac{1}{3} + \mu^2 = 1 \quad \mu^2 = \frac{1}{6} \quad \dots(3)$$

$$\gamma^2 + \delta^2 + \mu^2 = 1$$

$$\gamma^2 + \delta^2 = 1 - \frac{1}{6} = \frac{5}{6}$$

For Q:

$$x\vec{u} + y\vec{v} + z\vec{w} = \hat{j}$$

$$x\vec{u} \cdot \vec{u} + y\vec{v} \cdot \vec{u} + z\vec{w} \cdot \vec{u} = \vec{u} \cdot \hat{j}$$

$$x = \frac{1}{\sqrt{3}}$$

For R: 

$$\vec{u}, \vec{v}, \vec{w}$$ are mutually perpendicular unit vectors.

$$\vert{}\vec{u} \cdot (\vec{v} \times \vec{w})\vert{}$$

$$\begin{bmatrix} \vec{u} & \vec{v} & \vec{w} \end{bmatrix} = 1$$

For S: 

$$\vec{u} \times (\vec{v} \times \vec{w})$$

$$(\vec{u} \cdot \vec{w})\vec{v} - (\vec{u} \cdot \vec{v})\vec{w}$$

$$= 0 - 0 = 0$$

Solution :

Given,

$$\vec a\times\vec c=\vec b$$

Take dot product of $$\vec a$$ on both sides :

$$\vec a\cdot(\vec a\times\vec c)=\vec a\cdot\vec b$$

But,

$$\vec a\cdot(\vec a\times\vec c)=0$$

since a vector is always perpendicular to its cross product.

Therefore,

$$\vec a\cdot\vec b=0$$

Now,

$$\vec a=\hat i+\hat j+\hat k$$

and

$$\vec b=\hat i-\hat j+\hat k$$

Hence,

$$\vec a\cdot\vec b=(1)(1)+(1)(-1)+(1)(1)$$

$$=1-1+1$$

$$=1$$

But,

$$\vec a\cdot\vec b=1\neq 0$$

which contradicts the necessary condition

$$\vec a\cdot\vec b=0$$

for the existence of a vector $$\vec c$$ satisfying

$$\vec a\times\vec c=\vec b$$

Therefore, no such vector $$\vec c$$ exists.