Let $\overrightarrow{\mathrm{a}}=2 \hat{i}-3 \hat{j}+\hat{k}, \quad \overrightarrow{\mathrm{~b}}=3 \hat{i}+2 \hat{j}+5 \hat{k}$ and a vector $\overrightarrow{\mathrm{c}}$ be such that $(\vec{a}-\vec{c}) \times \vec{b}=-18 \hat{i}-3 \hat{j}+12 \hat{k}$ and $\vec{a} \cdot \vec{c}=3$. If $\vec{b} \times \vec{c}=\vec{d}$, then $|\vec{a} \cdot \vec{d}|$ is equal to :

<p>To solve the problem, we start with the given equation:</p> <p>$(\vec{a} - \vec{c}) \times \vec{b} = -18 \hat{i} - 3 \hat{j} + 12 \hat{k}.$</p> <p>Expanding this, we have:</p> <p>$$\vec{a} \times \vec{b} - \vec{c} \times \vec{b} = -18 \hat{i} - 3 \hat{j} + 12 \hat{k}.$$</p> <p>Let $\vec{d} = \vec{c} \times \vec{b}$. Substituting this into the equation, we get:</p> <p>$\vec{a} \times \vec{b} + \vec{d} = -18 \hat{i} - 3 \hat{j} + 12 \hat{k}.$</p> <p>Now, we take the dot product of both sides with $\vec{a}$:</p> <p>$$\vec{a} \cdot (\vec{a} \times \vec{b}) + \vec{a} \cdot \vec{d} = \vec{a} \cdot (-18 \hat{i} - 3 \hat{j} + 12 \hat{k}).$$</p> <p>Since the dot product $\vec{a} \cdot (\vec{a} \times \vec{b})$ is zero (the dot product of any vector with the cross product of itself with another vector is zero), we have:</p> <p>$\vec{a} \cdot \vec{d} = -36 + 9 + 12 = -15.$</p> <p>Thus, the magnitude of $\vec{a} \cdot \vec{d}$ is:</p> <p>$|\vec{a} \cdot \vec{d}| = 15.$</p>