If $$\vec{x} = 3\hat{i} - 6\hat{j} - \hat{k}$$, $$\vec{y} = \hat{i} + 4\hat{j} - 3\hat{k}$$ and $$\vec{z} = 3\hat{i} - 4\hat{j} - 12\hat{k}$$, then the magnitude of the projection of $$\vec{x} \times \vec{y}$$ on $$\vec{z}$$ is:

We are given three vectors:

$$\vec{x} = 3\hat{i} - 6\hat{j} - \hat{k}$$

$$\vec{y} = \hat{i} + 4\hat{j} - 3\hat{k}$$

$$\vec{z} = 3\hat{i} - 4\hat{j} - 12\hat{k}$$

We need to find the magnitude of the projection of $$\vec{x} \times \vec{y}$$ onto $$\vec{z}$$. The magnitude of the projection of a vector $$\vec{a}$$ onto a vector $$\vec{b}$$ is given by $$\frac{| \vec{a} \cdot \vec{b} |}{|\vec{b}|}$$. Here, $$\vec{a} = \vec{x} \times \vec{y}$$ and $$\vec{b} = \vec{z}$$, so the magnitude is $$\frac{| (\vec{x} \times \vec{y}) \cdot \vec{z} |}{|\vec{z}|}$$.

First, compute the cross product $$\vec{x} \times \vec{y}$$. The cross product is found using the determinant formula:

$$\vec{x} \times \vec{y} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -6 & -1 \\ 1 & 4 & -3 \end{vmatrix}$$

Expanding along the first row:

$$\hat{i} \begin{vmatrix} -6 & -1 \\ 4 & -3 \end{vmatrix} - \hat{j} \begin{vmatrix} 3 & -1 \\ 1 & -3 \end{vmatrix} + \hat{k} \begin{vmatrix} 3 & -6 \\ 1 & 4 \end{vmatrix}$$

Compute each 2x2 determinant:

• For $$\hat{i}$$: $$(-6)(-3) - (-1)(4) = 18 - (-4) = 18 + 4 = 22$$
• For $$\hat{j}$$: $$(3)(-3) - (-1)(1) = -9 - (-1) = -9 + 1 = -8$$, and with the negative sign from expansion, it becomes $$-(-8) = 8$$
• For $$\hat{k}$$: $$(3)(4) - (-6)(1) = 12 - (-6) = 12 + 6 = 18$$

So, $$\vec{x} \times \vec{y} = 22\hat{i} + 8\hat{j} + 18\hat{k}$$

Next, compute the dot product of $$\vec{x} \times \vec{y}$$ and $$\vec{z}$$:

$$(\vec{x} \times \vec{y}) \cdot \vec{z} = (22)(3) + (8)(-4) + (18)(-12)$$

Calculate step by step:

$$22 \times 3 = 66$$

$$8 \times (-4) = -32$$

$$18 \times (-12) = -216$$

Adding these: $$66 + (-32) + (-216) = 66 - 32 - 216 = 34 - 216 = -182$$

So, $$(\vec{x} \times \vec{y}) \cdot \vec{z} = -182$$

Now, find the magnitude of $$\vec{z}$$:

$$|\vec{z}| = \sqrt{(3)^2 + (-4)^2 + (-12)^2} = \sqrt{9 + 16 + 144} = \sqrt{169} = 13$$

The magnitude of the projection is:

$$\frac{| (\vec{x} \times \vec{y}) \cdot \vec{z} |}{|\vec{z}|} = \frac{|-182|}{13} = \frac{182}{13}$$

Dividing: $$182 \div 13 = 14$$, since $$13 \times 14 = 182$$.

Therefore, the magnitude is 14.

Hence, the correct answer is Option A.