Let <span class="math-tex">\(\vec{a}, \vec{b}\)</span> and <span class="math-tex">\(\vec{c}\)</span> be three non-zero vectors, no two of which are collinear. If the vector <span class="math-tex">\(\vec{a}+2\vec{b}\)</span> is collinear with <span class="math-tex">\(\vec{c}\)</span> and <span class="math-tex">\(\vec{b}+3\vec{c}\)</span> is collinear with <span class="math-tex">\(\vec{a}\)</span>, then <span class="math-tex">\(\vec{a} + 2\vec{b}+6\vec{c}\)</span> is equal to
Step-by-step Solution:
\[ \text{Let } a+2b = xc \quad \text{and} \quad b+3c = ya. \] \[ \text{Then } a+2b+6c = (x+6)c \] \[ \text{and also, } a+2b+6c = (1+2y)a. \] \[ \text{So } (x+6)c = (1+2y)a. \] \[ \text{Since } a, b, \text{ and } c \text{ are non-zero and non-collinear,} \] \[ \text{we have } x+6 = 0 \] \[ \text{and } 1+2y = 0, \text{ i.e. } x=-6 \text{ and } y=-\frac{1}{2}. \] \[ \text{In either case, we have } a+2b+6c = 0. \]