Let $$a, b$$ and $$c$$ be distinct positive numbers. If the vectors $$a\hat{i} + a\hat{j} + c\hat{k}$$, $$\hat{i} + \hat{k}$$ and $$c\hat{i} + c\hat{j} + b\hat{k}$$ are co-planar, then $$c$$ is equal to:

We are told that the three vectors

$$\vec{v_1}=a\hat i+a\hat j+c\hat k,\qquad \vec{v_2}=1\hat i+0\hat j+1\hat k,\qquad \vec{v_3}=c\hat i+c\hat j+b\hat k$$

are co-planar. Three vectors are co-planar precisely when their scalar triple product is zero. The scalar triple product formula is

$$[\vec{v_1}\;\vec{v_2}\;\vec{v_3}]= \begin{vmatrix} v_{1x}&v_{1y}&v_{1z}\\ v_{2x}&v_{2y}&v_{2z}\\ v_{3x}&v_{3y}&v_{3z} \end{vmatrix}=0.$$

Now we substitute the components of our vectors into the determinant:

$$ \begin{vmatrix} a & a & c \\ 1 & 0 & 1 \\ c & c & b \end{vmatrix}=0. $$

We expand this determinant along the first row. First, write the general expansion:

$$\left| \begin{array}{ccc} A_{11}&A_{12}&A_{13}\\ A_{21}&A_{22}&A_{23}\\ A_{31}&A_{32}&A_{33} \end{array}\right| =A_{11}(A_{22}A_{33}-A_{23}A_{32}) -A_{12}(A_{21}A_{33}-A_{23}A_{31}) +A_{13}(A_{21}A_{32}-A_{22}A_{31}).$$

Comparing with our entries, we identify

$$A_{11}=a,\;A_{12}=a,\;A_{13}=c,$$ $$A_{21}=1,\;A_{22}=0,\;A_{23}=1,$$ $$A_{31}=c,\;A_{32}=c,\;A_{33}=b.$$

Substituting, we get

$$ \begin{aligned} [\vec{v_1}\;\vec{v_2}\;\vec{v_3}]&= a\bigl(0\cdot b-1\cdot c\bigr) -a\bigl(1\cdot b-1\cdot c\bigr) +c\bigl(1\cdot c-0\cdot c\bigr)\\[4pt] &=a(0-bc)-a(b-c)+c(c-0)\\[4pt] &=-ac-a(b-c)+c^2\\[4pt] &=-ac-ab+ac+c^2\\[4pt] &=-ab+c^2. \end{aligned} $$

Because the three vectors are co-planar, the scalar triple product is zero, so we set

$$-ab+c^2=0.$$

Rearranging, we obtain

$$c^2=ab.$$

All three of $$a,b,c$$ are positive, and the distinctness condition merely tells us they are not equal; hence we take the positive square root:

$$c=\sqrt{ab}.$$

Hence, the correct answer is Option D.