If the point $$(2, \alpha, \beta)$$ lies on the plane which passes through the points (3,4,2) and (7,0,6) and is perpendicular to the plane $$2x - 5y = 15$$, then $$2\alpha - 3\beta$$ is equal to:

We have to find the value of $$2\alpha-3\beta$$ for the point $$(2,\alpha,\beta)$$ which lies on a certain plane. This required plane passes through the two fixed points $$P(3,4,2)$$ and $$Q(7,0,6)$$ and is given to be perpendicular to the plane $$2x-5y=15$$.

The plane $$2x-5y=15$$ can be written in normal form $$\vec n_1\cdot\vec r=d$$, where its normal vector is obtained directly from the coefficients of $$x,y,z$$. So we take

$$\vec n_1=\langle 2,-5,0\rangle.$$

For our required plane, every pair of points on it gives a direction vector that lies entirely within the plane. Let us form two such direction vectors:

From $$P(3,4,2)$$ to $$Q(7,0,6)$$ we get $$\vec{PQ}=\langle 7-3,\;0-4,\;6-2\rangle =\langle 4,-4,4\rangle.$$

From $$P(3,4,2)$$ to $$R(2,\alpha,\beta)$$ we get $$\vec{PR}=\langle 2-3,\;\alpha-4,\;\beta-2\rangle =\langle -1,\;\alpha-4,\;\beta-2\rangle.$$

A normal vector to the required plane is given by the cross-product of any two non-parallel direction vectors lying in that plane. Using the formula $$\vec a\times\vec b=\begin{vmatrix} \hat i & \hat j & \hat k\\ a_1 & a_2 & a_3\\ b_1 & b_2 & b_3 \end{vmatrix},$$ we compute $$\vec n_2=\vec{PQ}\times\vec{PR}.$$

Writing out the determinant step by step:

$$ \vec n_2= \hat i\bigl((-4)(\beta-2)-4(\alpha-4)\bigr) -\hat j\bigl(4(\beta-2)-4(-1)\bigr) +\hat k\bigl(4(\alpha-4)-(-4)(-1)\bigr). $$

Now we simplify each component one by one.

For the $$\hat i$$ component: $$(-4)(\beta-2)-4(\alpha-4) =-4\beta+8-4\alpha+16 =-4\alpha-4\beta+24.$$

For the $$\hat j$$ component (remember the overall minus sign in front of $$\hat j$$): $$4(\beta-2)-4(-1)=4\beta-8+4 =4\beta-4,$$ so the actual $$\hat j$$ component becomes $$-(4\beta-4)=-4\beta+4.$$

For the $$\hat k$$ component: $$4(\alpha-4)-(-4)(-1)=4\alpha-16-4=4\alpha-20.$$

Thus $$\vec n_2=\langle -4\alpha-4\beta+24,\;-4\beta+4,\;4\alpha-20\rangle.$$ We can factor out a convenient common factor of $$4$$ to keep the numbers small:

$$\vec n_2=4\langle -\alpha-\beta+6,\;-\beta+1,\;\alpha-5\rangle.$$

Since any non-zero scalar multiple of a normal vector is also a normal vector, it is simpler to use

$$\vec v=\langle -\alpha-\beta+6,\;-\beta+1,\;\alpha-5\rangle$$

as the normal vector of our required plane.

The problem states that this required plane is perpendicular to the plane $$2x-5y=15$$. Two planes are perpendicular precisely when their normal vectors are perpendicular. In vector form this condition is expressed via the dot product formula

$$\vec v\cdot\vec n_1=0.$$

Substituting $$\vec v=\langle -\alpha-\beta+6,\;-\beta+1,\;\alpha-5\rangle$$ and $$\vec n_1=\langle 2,-5,0\rangle$$, we obtain

$$ \langle -\alpha-\beta+6,\;-\beta+1,\;\alpha-5\rangle\cdot \langle 2,-5,0\rangle=0. $$

Now we take the dot product term by term:

$$ 2(-\alpha-\beta+6)+(-5)(-\beta+1)+0(\alpha-5)=0. $$

Simplifying each term:

First term: $$2(-\alpha-\beta+6)=-2\alpha-2\beta+12.$$

Second term: $$(-5)(-\beta+1)=5\beta-5.$$

The third term is zero because it is multiplied by zero.

Adding these contributions: $$(-2\alpha-2\beta+12)+(5\beta-5)=0.$$ Combine like terms in $$\beta$$ and in the constants:

$$-2\alpha+(-2\beta+5\beta)+(12-5)=0,$$ which gives $$-2\alpha+3\beta+7=0.$$

Moving everything except the $$2\alpha$$ term to the right side:

$$2\alpha=3\beta+7.$$

Therefore

$$2\alpha-3\beta=7.$$

Hence, the correct answer is Option B.