The plane containing the line
\(\vec { r } =(\hat { i } -\hat { j } +3\hat { k } )+t(2\hat { i } -\hat { j } +4\hat { k } )\)
∴ The required plane is passing through the point \(\vec { a } =\hat { i } -\hat { j } +3\hat { k } \) and parallel to a vector \(\vec { b } =2\hat { i } -\hat { j } +4\hat { k } \) Also, the plane is perpendicular to the plane 
\(\vec { c } =\hat { i } +2\hat { j } +\hat { k } \) 
∴ The parametric form of vector equation of the plane passing through one point (\(\vec { a } \)) and parallel to two vectors \(\vec { b } \) and \(\vec { c } \) 
\(\vec { r } =\vec { a } +s\vec { b } +t\vec { c } \) where s, t ∈ R
⇒ \(\vec { r } .(\hat { i } -\hat { j } +3\hat { k } )+s(2\hat { i } -\hat { j } +4\hat { k } )+t(\hat { i } +2\hat { j } +\hat { k } )\) s, t  ∈ R 
Cartesian equation is
\(\left| \begin{matrix} x-{ x }_{ 1 } & y-{ y }_{ 1 } & z-{ z }_{ 1 } \\ { b }_{ 1 } & { b }_{ 2 } & { b }_{ 3 } \\ { c }_{ 1 } & { c }_{ 2 } & { c }_{ 3 } \end{matrix} \right| =0\)
\(\Rightarrow \left| \begin{matrix} x-1 & y+1 & z-3 \\ 2 & -1 & 4 \\ 1 & 2 & 1 \end{matrix} \right| =0\)
⇒ (x - 1)(-1 - 8) - (y + 1)(2 - 4) + (z - 3)(4 + 1) = 0
⇒ (x - 1)(-9) - (y + 1)(-2) + (z - 3)5 = 0
⇒ -9x + 9 +2y + 2 + 5z - 15 = 0
⇒ -9x + 2y + 5z - 4 = 0
⇒ 9x - 2y - 5z + 4 = 0

By definition, \(\vec { a } \times \vec { b } \) \(=\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ -2 & 3 & -2 \\ 3 & -1 & 3 \end{matrix} \right| =7\hat { i } -7\hat { k } \)
Then, \((\vec { a } \times \vec { b } )\times \vec { c } =\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ 7 & 0 & -7 \\ 2 & -5 & 1 \end{matrix} \right| =-35\hat { i } -21\hat { j } -35\hat { k } \)......(1)
\(\vec { b } \times \vec { c } =\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ 3 & -1 & 3 \\ 2 & -5 & 1 \end{matrix} \right| =14\hat { i } +3\hat { j } -13\hat { k } \)
\(\vec { a } (\vec { b } \times \vec { c } )=\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ -2 & 3 & -2 \\ 14 & 3 & -13 \end{matrix} \right| =-33\hat { i } -54\hat { j } -48\hat { k } \)....(2)
Therefore, equations (1) and (2) show that \((\vec { a } \times \vec { b } )\times \vec { c } \)\(​​\neq \)\((\vec { a } \times \vec { b } )\times \vec { c } \)
 

Let A be the origin, \(\vec { b } \) be the position vector of B and \(\vec {c } \) be the position vector of C .
Now D is the midpoint of BC , and so the position vector of D \(\frac{\vec{b}+\vec{c}}{2}\). There, we get

\({ \left| \vec { AD } \right| }^{ 2 }=\vec { AD } .\vec { AD } \)= \(\left( \frac { \vec { b } +\vec { c } }{ 2 } \right) .\left( \frac { \vec { b } +\vec { c } }{ 2 } \right) \)= \(\frac { 1 }{ 4 } ({ \left| \vec { b } \right| }^{ 2 }+{ \left| \vec { c } \right| }^{ 2 }+2\overrightarrow { b } .\vec { c } )\)....(1)
Now, \(\vec { BD } =\vec { AD } -\vec { AB } \) = \(\frac { \vec { b } +\vec { c } }{ 2 } -\vec { b }=\frac { \vec {c } -\vec { b} }{ 2 }\)
Then, we get, =\({ \left| \vec { BD } \right| }^{ 2 }=\vec { BD } .\vec {BD } \) = \(\left( \frac { \vec { c } -\vec { b} }{ 2 } \right) .\left( \frac { \vec { c } -\vec { b} }{ 2 } \right) \) = \(\frac { 1 }{ 4 } ({ \left| \vec { b } \right| }^{ 2 }+{ \left| \vec { c } \right| }^{ 2 }-2\overrightarrow { b } .\vec { c } )\)....(2)
Now, adding (1) and (2), we get
Therefore, \({ \left| \vec { AD } \right| }^{ 2 }+{ \left| \vec { BD } \right| }^{ 2 }=\frac { 1 }{ 4 } ({ \left| \vec { b } \right| }^{ 2 }+{ \left| \vec { c } \right| }^{ 2 }+2\overrightarrow { b } .\vec { c } )+\frac { 1 }{ 4 } ({ \left| \vec { b } \right| }^{ 2 }+{ \left| \vec { c } \right| }^{ 2 }-2\overrightarrow { b } .\vec { c } )=\frac { 1 }{ 2 } ({ \left| \vec { b } \right| }^{ 2 }+{ \left| \vec { c } \right| }^{ 2 })\)
⇒ \({ \left| \vec { AD } \right| }^{ 2 }+{ \left| \vec { BD } \right| }^{ 2 }=\frac { 1 }{ 2 } ({ \left| \vec { AB } \right| }^{ 2 }+{ \left| \vec { AC } \right| }^{ 2 })\)
Hence, \({ \left| \vec { AB } \right| }^{ 2 }+{ \left| \vec { AC } \right| }^{ 2 }=2({ \left| \vec { AD } \right| }^{ 2 }+{ \left| \vec { BD } \right| }^{ 2 })\)

\(\hat { d } =\frac { 3\hat { i } -4\hat { j } +5\hat { k } }{ \sqrt { { 3 }^{ 2 }+(-4)+{ 5 }^{ 2 } } } =\frac { 3\hat { i } -4\hat { j } +5\hat { k } }{ \sqrt { 9+16+25 } } \)
\(=\frac { 3\hat { i } -4\hat { j } +5\hat { k } }{ \sqrt { 50 } } =\frac { 3\hat { i } -4\hat { j } +5\hat { k } }{ 5\sqrt { 2 } } \)
[∵ 3, -4, 5 are direction ratios]. The equation of the plane at a distance p from the origin and perpendicular to the unit normal vector \(\hat { d }\ is\ \vec { r } .\hat { d } =p\)
Equation of the required plane is
\(\vec { r } .\left( \frac { 3\hat { i } -4\hat { j } +5\hat { k } }{ 5\sqrt { 2 } } \right) =7\)

Given \(\vec { F } =3\hat { i } +4\hat { j } -5\hat { k } \)
\(\vec { r } \) = (Force acting through the point) - (force acting to the point)
= \((4\hat { i } +2\hat { j } -3\hat { k } )-(2\hat { i } -\hat { j } +4\hat { k } )\)
= \(2\hat { i } +5\hat { j } -7\hat { k } \)
Torque = \(\vec { c } =\hat { r } \times \hat { F } \)
=  \(\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ 2 & 5 & -7 \\ 3 & 4 & -5 \end{matrix} \right| =\hat { i } \left| \begin{matrix} 5 & -7 \\ 4 & -5 \end{matrix} \right| -\hat { j } \left| \begin{matrix} 2 & -7 \\ 3 & -5 \end{matrix} \right| +\hat { k } \left| \begin{matrix} 2 & 5 \\ 3 & 4 \end{matrix} \right| \)
\(\hat { i } =\hat { i } (-25+28)-\hat { j } (-10+21)+\hat { j } (8-15)\)
= \(3\hat { i } -11\hat { j } -7\hat { k } \)
∴ MagnitudeoftheTorque = \(\sqrt { { 3 }^{ 2 }+(-11)^{ 2 }+(-7)^{ 2 } } \)
= \(\sqrt { 9+121+49 } =\sqrt { 179 } \)
Hence, the direction cosines are \(\left( \frac { 3 }{ \sqrt { 179 } } ,\frac { -11 }{ \sqrt { 179 } } ,\frac { -7 }{ \sqrt { 179 } } \right) \).

Let ABC be an isosceles triangle with AB = AC and let AD is the median 
D is mid-point of BC.
\( \overrightarrow{A D}=\frac{1}{2}(\overrightarrow{A B}+\overrightarrow{A C}) \)
\( \overrightarrow{B C}=\overrightarrow{B A}+\overrightarrow{A C} \)
\( \overline{D A} \cdot \overrightarrow{D B}=-\overrightarrow{A D} \cdot\left(\frac{-1}{2} \overrightarrow{C B}\right) \)
\(=-\overrightarrow{A D} \cdot\left(\frac{1}{2} \overrightarrow{B C}\right) \)
\(=\frac{1}{2} \overrightarrow{A D} \cdot \overrightarrow{B C} \)
\(=\frac{1}{4}(\overrightarrow{A B}+\overrightarrow{A C}) \cdot(\overrightarrow{B A}+\overrightarrow{A C}) \)
\(=\frac{1}{4}(\overrightarrow{A B}+\overrightarrow{A C}) \cdot(\overrightarrow{A C}-\overrightarrow{A B}) \)
\(=\frac{1}{4}[(\overrightarrow{A C} \cdot \overrightarrow{A C})-(\overrightarrow{A B} \cdot \overrightarrow{A B})] \)
\(=\frac{1}{4}\left(A C^{2}-A B^{2}\right) \)
\(=\frac{1}{4}(0)=0 \)
\(\overrightarrow{D A} \cdot \overrightarrow{D B}=0\)
\(\overrightarrow{D A} \perp \overrightarrow{D B}\)

We observe that the straight line \(\frac { x-1 }{ 4 } =\frac { 2-y }{ 6 } =\frac { z-4 }{ 12 } \) is parallel to the vector \(4\hat { i } -6\hat { j } +12\hat { k } \) and the straight line \(\frac { x-3 }{ -2 } =\frac { y-3 }{ 3 } =\frac { 5-z }{ 6 } \) is parallel to the vector \(2\hat { i } +3\hat { j } -6\hat { k } \)
Since \(4\hat { i } -6\hat { j } +12\hat { k } =-2(-2\hat { i } +3\hat { j } -6\hat { k } )\), two vectors are parallel, and hence the two straight lines are parallel.

Let \(\vec { a } =-6\hat { i } +14\hat { j } +10\hat { k } \), \(\vec { b } =14\hat { i } -10\hat { j } -6\hat { k } \) and \(\vec { c } =2\hat { i } +4\hat { j } -2\hat { k } \)
Volume of the parallelepiped having \(\vec { a } ,\vec { b } \) and \(\vec { c } \) as its co-terminus edges is \(\vec { a } .(\vec { b } \times \vec { c } )\).
∴ \(\vec { a } .(\vec { b } \times \vec { c } )=\left| \begin{matrix} -6 & 14 & 10 \\ 14 & -10 & -6 \\ 2 & 4 & -2 \end{matrix} \right| \)
= \(-6\left| \begin{matrix} -10 & -6 \\ 4 & -2 \end{matrix} \right| -14\left| \begin{matrix} 14 & -6 \\ 2 & -2 \end{matrix} \right| +10\left| \begin{matrix} 14 & -10 \\ 2 & 4 \end{matrix} \right| \)
= -6(20 + 24) - 14(-28 + 12) + 10(56 + 20)
= -6(44) -14(-16) + 10(76)
= -264 + 224 + 760 = 720.
∴ Volume of the required parallelepiped = 720 cubic units.