Let a = 4 i - j + 3 k , b = 10 i + 2 j - k and a vector c be such...

Given $2(\vec{a}\times\vec{b}) + 3(\vec{b}\times\vec{c}) = \vec{0}$

$\Rightarrow 2(\vec{a}\times\vec{b}) - 3(\vec{c}\times\vec{b}) = \vec{0}$

$\Rightarrow (2\vec{a} - 3\vec{c}) \times \vec{b} = \vec{0}$

Since the cross product is zero, the vectors are collinear:

$2\vec{a} - 3\vec{c} = \lambda \vec{b}$

$\Rightarrow 3\vec{c} = 2\vec{a} - \lambda \vec{b}$

Taking the dot product with $\vec{a}$ on both sides:

$3(\vec{a}\cdot\vec{c}) = 2|\vec{a}|^2 - \lambda(\vec{a}\cdot\vec{b})$

We have $\vec{a} = 4\hat{i} - \hat{j} + 3\hat{k}$ and $\vec{b} = 10\hat{i} + 2\hat{j} - \hat{k}$.

$|\vec{a}|^2 = 4^2 + (-1)^2 + 3^2 = 26$

$\vec{a}\cdot\vec{b} = 4(10) + (-1)(2) + 3(-1) = 40 - 2 - 3 = 35$

Given $\vec{a}\cdot\vec{c} = 15$, substituting these values:

$3(15) = 2(26) - 35\lambda$

$45 = 52 - 35\lambda$

$35\lambda = 7 \Rightarrow \lambda = \dfrac{1}{5}$

Substituting $\lambda$ back into the equation for $\vec{c}$:

$3\vec{c} = 2\vec{a} - \dfrac{1}{5}\vec{b}$

We need to find $\vec{c}\cdot(\hat{i}+\hat{j}-3\hat{k})$. Let $\vec{v} = \hat{i}+\hat{j}-3\hat{k}$.

Taking the dot product with $\vec{v}$ on both sides:

$3(\vec{c}\cdot\vec{v}) = 2(\vec{a}\cdot\vec{v}) - \dfrac{1}{5}(\vec{b}\cdot\vec{v})$

Calculating $\vec{a}\cdot\vec{v}$ and $\vec{b}\cdot\vec{v}$:

$\vec{a}\cdot\vec{v} = 4(1) + (-1)(1) + 3(-3) = 4 - 1 - 9 = -6$

$\vec{b}\cdot\vec{v} = 10(1) + 2(1) + (-1)(-3) = 10 + 2 + 3 = 15$

Substituting these values:

$3(\vec{c}\cdot\vec{v}) = 2(-6) - \dfrac{1}{5}(15)$

$3(\vec{c}\cdot\vec{v}) = -12 - 3 = -15$

$\vec{c}\cdot\vec{v} = -5$

Answer: $-5$