Let $$\vec{a} = 2\hat{i} - 3\hat{j} + \hat{k}$$, $$\vec{b} = 3\hat{i} + 2\hat{j} + 5\hat{k}$$ and a vector $$\vec{c}$$ 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 :

Let $$\vec{a}=2\hat{i}-3\hat{j}+\hat{k}$$, $$\vec{b}=3\hat{i}+2\hat{j}+5\hat{k}$$ and $$\vec{c}=x\hat{i}+y\hat{j}+z\hat{k}$$.

The condition $$(\vec{a}-\vec{c})\times\vec{b}=-18\hat{i}-3\hat{j}+12\hat{k}$$ gives three scalar equations.
Write $$\vec{a}-\vec{c}=(2-x)\hat{i}+(-3-y)\hat{j}+(1-z)\hat{k}.$$

For two vectors $$\vec{p}=(p_1,p_2,p_3)$$ and $$\vec{q}=(q_1,q_2,q_3)$$, the cross product is
$$\vec{p}\times\vec{q}=\bigl(p_2q_3-p_3q_2\bigr)\hat{i}-\bigl(p_1q_3-p_3q_1\bigr)\hat{j}+\bigl(p_1q_2-p_2q_1\bigr)\hat{k}.$$

Putting $$\vec{p}=(2-x,-3-y,1-z)$$ and $$\vec{q}=(3,2,5)$$:

$$\begin{aligned} (\vec{a}-\vec{c})\times\vec{b} &=\bigl[5(-3-y)-2(1-z)\bigr]\hat{i}\\ &\;\;-\bigl[5(2-x)-3(1-z)\bigr]\hat{j}\\ &\;\;+\bigl[2(2-x)-3(-3-y)\bigr]\hat{k}. \end{aligned}$$

Equating with $$-18\hat{i}-3\hat{j}+12\hat{k}$$ gives

$$5(-3-y)-2(1-z)=-18$$ $$-(1)$$
$$5(2-x)-3(1-z)=3$$ $$-(2)$$
$$2(2-x)-3(-3-y)=12$$ $$-(3)$$

Simplify each:

From $$(1):\; -15-5y-2+2z=-18 \;\Rightarrow\; 5y-2z=1 \;-(4)$$

From $$(2):\; 10-5x-3+3z=3 \;\Rightarrow\; 5x-3z=4 \;-(5)$$

From $$(3):\; 4-2x+9+3y=12 \;\Rightarrow\; 2x-3y=1 \;-(6)$$

The second given condition is $$\vec{a}\cdot\vec{c}=3$$:
$$2x-3y+z=3 \;-(7)$$

Solve equations $$(4)-(7).$$

From $$(6):\; 2x-3y=1 \;\Rightarrow\; y=\frac{2x-1}{3}.$$ Substitute this in $$(7):$$
$$2x-3\!\left(\frac{2x-1}{3}\right)+z=3$$
$$\Rightarrow\;2x-(2x-1)+z=3 \;\Rightarrow\;1+z=3 \;\Rightarrow\; z=2.$$

Insert $$z=2$$ in $$(5):$$
$$5x-3(2)=4 \;\Rightarrow\; 5x-6=4 \;\Rightarrow\; x=2.$$

Put $$x=2$$ in $$(6):$$
$$2(2)-3y=1 \;\Rightarrow\; 4-3y=1 \;\Rightarrow\; y=1.$$

Hence $$\vec{c}=2\hat{i}+\hat{j}+2\hat{k}.$$

Now find $$\vec{d}=\vec{b}\times\vec{c}.$$ Compute the determinant:
$$\vec{d}= \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ 3 & 2 & 5\\ 2 & 1 & 2 \end{vmatrix} = \bigl(2\cdot2-5\cdot1\bigr)\hat{i}-\bigl(3\cdot2-5\cdot2\bigr)\hat{j}+\bigl(3\cdot1-2\cdot2\bigr)\hat{k}.$$

Simplifying:
$$\vec{d}=-1\hat{i}+4\hat{j}-1\hat{k}= -\hat{i}+4\hat{j}-\hat{k}.$$

Finally compute $$\vec{a}\cdot\vec{d}:$$
$$\vec{a}\cdot\vec{d}=(2,-3,1)\cdot(-1,4,-1)=-2-12-1=-15.$$ Therefore
$$\lvert\vec{a}\cdot\vec{d}\rvert=15.$$

Option D is correct.