Question:

Step 1: Simplify the first vector equation using properties of the cross product. We are given the equation $(\vec{a}+\vec{b})\times\vec{c}=\vec{c}\times(-2\vec{a}+3\vec{b})$. Using the property $\vec{X} \times \vec{Y} = -(\vec{Y} \times \vec{X})$, we can rewrite the right side: $$ (\vec{a}+\vec{b})\times\vec{c} = -((-2\vec{a}+3\vec{b})\times\vec{c}) $$ Move all terms to one side: $$ (\vec{a}+\vec{b})\times\vec{c} + (-2\vec{a}+3\vec{b})\times\vec{c} = \vec{0} $$ Using the distributive property of the cross product, $(\vec{X}+\vec{Y})\times\vec{Z} = \vec{X}\times\vec{Z} + \vec{Y}\times\vec{Z}$: $$ ((\vec{a}+\vec{b}) + (-2\vec{a}+3\vec{b}))\times\vec{c} = \vec{0} $$ Combine the vector terms inside the parenthesis: $$ (\vec{a}+\vec{b}-2\vec{a}+3\vec{b})\times\vec{c} = \vec{0} $$ $$ (-\vec{a}+4\vec{b})\times\vec{c} = \vec{0} $$ This implies that vector $\vec{c}$ is parallel to the vector $(-\vec{a}+4\vec{b})$. Therefore, we can write $\vec{c}$ as a scalar multiple of $(-\vec{a}+4\vec{b})$: $$ \vec{c} = \lambda (-\vec{a}+4\vec{b}) $$ for some scalar $\lambda$. Step 2: Calculate the vector $(-\vec{a}+4\vec{b})$. Given $\vec{a}=4\hat{i}-\hat{j}+\hat{k}$ and $\vec{b}=11\hat{i}-\hat{j}+\hat{k}$. First, calculate $-\vec{a}$: $$ -\vec{a} = -(4\hat{i}-\hat{j}+\hat{k}) = -4\hat{i}+\hat{j}-\hat{k} $$ Next, calculate $4\vec{b}$: $$ 4\vec{b} = 4(11\hat{i}-\hat{j}+\hat{k}) = 44\hat{i}-4\hat{j}+4\hat{k} $$ Now, add these two vectors to find $(-\vec{a}+4\vec{b})$: $$ -\vec{a}+4\vec{b} = (-4\hat{i}+\hat{j}-\hat{k}) + (44\hat{i}-4\hat{j}+4\hat{k}) $$ $$ -\vec{a}+4\vec{b} = (-4+44)\hat{i}+(1-4)\hat{j}+(-1+4)\hat{k} $$ $$ -\vec{a}+4\vec{b} = 40\hat{i}-3\hat{j}+3\hat{k} $$ So, we have: $$ \vec{c} = \lambda (40\hat{i}-3\hat{j}+3\hat{k}) $$ Step 3: Calculate the vector $(2\vec{a}+3\vec{b})$. This vector is needed for the second given equation. First, calculate $2\vec{a}$: $$ 2\vec{a} = 2(4\hat{i}-\hat{j}+\hat{k}) = 8\hat{i}-2\hat{j}+2\hat{k} $$ Next, calculate $3\vec{b}$: $$ 3\vec{b} = 3(11\hat{i}-\hat{j}+\hat{k}) = 33\hat{i}-3\hat{j}+3\hat{k} $$ Now, add these two vectors to find $(2\vec{a}+3\vec{b})$: $$ 2\vec{a}+3\vec{b} = (8\hat{i}-2\hat{j}+2\hat{k}) + (33\hat{i}-3\hat{j}+3\hat{k}) $$ $$ 2\vec{a}+3\vec{b} = (8+33)\hat{i}+(-2-3)\hat{j}+(2+3)\hat{k} $$ $$ 2\vec{a}+3\vec{b} = 41\hat{i}-5\hat{j}+5\hat{k} $$ Step 4: Use the second given equation to find the scalar $\lambda$. We are given $(2\vec{a}+3\vec{b})\cdot\vec{c}=1670$. Substitute the expressions for $(2\vec{a}+3\vec{b})$ and $\vec{c}$: $$ (41\hat{i}-5\hat{j}+5\hat{k})\cdot (\lambda (40\hat{i}-3\hat{j}+3\hat{k})) = 1670 $$ Factor out $\lambda$: $$ \lambda ((41\hat{i}-5\hat{j}+5\hat{k})\cdot (40\hat{i}-3\hat{j}+3\hat{k})) = 1670 $$ Perform the dot product: $$ \lambda ((41)(40) + (-5)(-3) + (5)(3)) = 1670 $$ $$ \lambda (1640 + 15 + 15) = 1670 $$ $$ \lambda (1670) = 1670 $$ Divide by 1670 to find $\lambda$: $$ \lambda = 1 $$ Step 5: Determine the vector $\vec{c}$. Substitute the value of $\lambda=1$ back into the expression for $\vec{c}$: $$ \vec{c} = 1 \cdot (40\hat{i}-3\hat{j}+3\hat{k}) $$ $$ \vec{c} = 40\hat{i}-3\hat{j}+3\hat{k} $$ Step 6: Calculate the square of the magnitude of $\vec{c}$, i.e., $|\vec{c}|^2$. For a vector $\vec{v} = x\hat{i}+y\hat{j}+z\hat{k}$, its magnitude squared is $|\vec{v}|^2 = x^2+y^2+z^2$. Using $\vec{c} = 40\hat{i}-3\hat{j}+3\hat{k}$: $$ |\vec{c}|^2 = (40)^2 + (-3)^2 + (3)^2 $$ $$ |\vec{c}|^2 = 1600 + 9 + 9 $$ $$ |\vec{c}|^2 = 1618 $$ The final answer is $\boxed{1618}$.