The area (in sq. units) of the parallelogram whose diagonals are along the vectors $$8\hat{i} - 6\hat{j}$$ and $$3\hat{i} + 4\hat{j} - 12\hat{k}$$, is:

We are given the two diagonal vectors of a parallelogram as $$\vec{d_1}=8\hat{i}-6\hat{j}$$ and $$\vec{d_2}=3\hat{i}+4\hat{j}-12\hat{k}.$$

For any parallelogram whose sides are $$\vec{a}$$ and $$\vec{b},$$ the diagonals become $$\vec{d_1}=\vec{a}+\vec{b}$$ and $$\vec{d_2}=\vec{a}-\vec{b}.$$

Using these, the vector cross-product of the diagonals is

$$\vec{d_1}\times\vec{d_2}=(\vec{a}+\vec{b})\times(\vec{a}-\vec{b}).$$

Expanding with the distributive property of the cross product, we have

$$\vec{d_1}\times\vec{d_2}=\vec{a}\times\vec{a}-\vec{a}\times\vec{b}+\vec{b}\times\vec{a}-\vec{b}\times\vec{b}.$$

Because any vector crossed with itself is the zero vector, $$\vec{a}\times\vec{a}=0$$ and $$\vec{b}\times\vec{b}=0.$$ Also, the cross product is anti-commutative, so $$\vec{b}\times\vec{a}=-(\vec{a}\times\vec{b}).$$ Substituting these facts, we get

$$\vec{d_1}\times\vec{d_2}=-\vec{a}\times\vec{b}-\vec{a}\times\vec{b}=-2\,\vec{a}\times\vec{b}.$$

Taking magnitudes on both sides gives

$$\lvert\vec{d_1}\times\vec{d_2}\rvert=2\,\lvert\vec{a}\times\vec{b}\rvert.$$

The magnitude $$\lvert\vec{a}\times\vec{b}\rvert$$ is precisely the area of the parallelogram. Therefore the required area is obtained from the diagonals by the formula

$$A=\frac{1}{2}\,\lvert\vec{d_1}\times\vec{d_2}\rvert.$$

Now we compute the cross product of the given diagonals. Writing the determinant,

$$ \vec{d_1}\times\vec{d_2}= \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ 8 & -6 & 0\\ 3 & 4 & -12 \end{vmatrix}. $$

Expanding along the first row, we have

$$ \vec{d_1}\times\vec{d_2}= \hat{i}\Bigl((-6)(-12)-0\cdot4\Bigr) - \hat{j}\Bigl(8(-12)-0\cdot3\Bigr) + \hat{k}\Bigl(8\cdot4-(-6)\cdot3\Bigr). $$

Simplifying each component step by step:

For $$\hat{i}$$ component: $$(-6)(-12)-0\cdot4=72-0=72.$$

For $$\hat{j}$$ component: $$8(-12)-0\cdot3=-96-0=-96,$$ and the overall sign in front of $$\hat{j}$$ makes it $$-\,-96=+96.$$

For $$\hat{k}$$ component: $$8\cdot4-(-6)\cdot3=32+18=50.$$

So we obtain

$$\vec{d_1}\times\vec{d_2}=72\hat{i}+96\hat{j}+50\hat{k}.$$

Next, we find its magnitude:

$$ \lvert\vec{d_1}\times\vec{d_2}\rvert =\sqrt{72^{2}+96^{2}+50^{2}} =\sqrt{5184+9216+2500} =\sqrt{16900} =130. $$

Finally, substituting into the area formula,

$$ A=\frac{1}{2}\times130=65\text{ square units}. $$

Hence, the correct answer is Option B.