Concept:

• The cross product of vector to itself = 0
• The cross product of collinear vectors = 0
• The dot product of collinear vectors = Product of their Magnitudes
• \(\rm \vec a \times \vec b=-\vec b \times \vec a\)
• For dot product \(\rm (\vec P + \vec Q) \cdot \vec R = (\vec P\cdot \vec R) +(\vec Q\cdot \vec R)\)
• For cross product \(\rm (\vec P + \vec Q) ×\vec R = (\vec P×\vec R) + (\vec Q×\vec R) \) 
• The unit vector in the direction of a \(\rm \vec P= \hat P={\vec P\over\left|\vec P\right|}\) 
• A vector \(\rm \vec X\) in direction of \(\rm \vec P\) = (Magnitude of \(\rm \vec X\)) × \(\rm \hat P\)

Calculation:

Given:

\(\rm\vec a × (2\hat i + 3\hat j + 4\hat k) = (2\hat i + 3\hat j + 4\hat k) × \vec b\)

⇒ \(\rm\vec a × (2\hat i + 3\hat j + 4\hat k) - (2\hat i + 3\hat j + 4\hat k) × \vec b=0\)

⇒ \(\rm\vec a × (2\hat i + 3\hat j + 4\hat k) + \vec b × (2\hat i + 3\hat j + 4\hat k) =0\)

⇒ \(\boldsymbol{\rm(\vec a + \vec b) × (2\hat i + 3\hat j + 4\hat k) =0}\)

It means the \(\rm\vec a + \vec b\) and \(\rm2\hat i + 3\hat j + 4\hat k\) are collinear vectors.

∴ \(\rm\vec a + \vec b = \left|\vec a + \vec b\right| × {2\hat i + 3\hat j + 4\hat k\over \left|2\hat i + 3\hat j + 4\hat k\right|}\) 

⇒ \(\rm\vec a + \vec b = \sqrt{29}× {2\hat i + 3\hat j + 4\hat k\over\sqrt{29}}\) 

⇒ \(\boldsymbol{\rm\vec a + \vec b = 2\hat i + 3\hat j + 4\hat k}\)

\(\rm (\vec a + \vec b)\cdot(-7\hat i + 2\hat j + 3\hat k) = \left(2\hat i + 3\hat j + 4\hat k\right)\cdot\left(-7\hat i + 2\hat j + 3\hat k\right)\)

⇒ \(\rm (\vec a + \vec b)\cdot(-7\hat i + 2\hat j + 3\hat k) = \left(2\times(-7) + 3\times2 + 4\times3\right)\) 

⇒ \(\rm (\vec a + \vec b)\cdot(-7\hat i + 2\hat j + 3\hat k) = \left(-14 +6 + 12\right)\) 

⇒ \(\boldsymbol{\rm (\vec a + \vec b)\cdot(-7\hat i + 2\hat j + 3\hat k) = 4}\)