Let a line L passing through the point $$(1, 1, 1)$$ be perpendicular to both the vectors $$2\hat{i} + 2\hat{j} + \hat{k}$$ and $$\hat{i} + 2\hat{j} + 2\hat{k}$$. If P(a, b, c) is the foot of perpendicular from the origin on the line L, then the value of $$34(a + b + c)$$ is :

The given vectors are $$\mathbf{v}_1 = 2\hat i + 2\hat j + \hat k$$ and $$\mathbf{v}_2 = \hat i + 2\hat j + 2\hat k$$. A line that is perpendicular to both $$\mathbf{v}_1$$ and $$\mathbf{v}_2$$ must have a direction vector equal to their cross-product.

Compute the cross-product: $$\mathbf{d}= \mathbf{v}_1 \times \mathbf{v}_2 = \begin{vmatrix} \hat i & \hat j & \hat k\\ 2 & 2 & 1\\ 1 & 2 & 2 \end{vmatrix} = \hat i(2\cdot2 - 1\cdot2) - \hat j(2\cdot2 - 1\cdot1) + \hat k(2\cdot2 - 2\cdot1) = 2\hat i - 3\hat j + 2\hat k.$$

Thus the required line $$L$$ passing through $$(1,1,1)$$ is $$(x,y,z) = (1,1,1) + t(2,-3,2).$$

Let $$P(a,b,c)$$ be the foot of the perpendicular from the origin to this line. For some parameter $$t=t_0$$, $$P = (1+2t_0,\; 1-3t_0,\; 1+2t_0).$$

The vector $$\overrightarrow{OP} = (1+2t_0,\; 1-3t_0,\; 1+2t_0)$$ must be perpendicular to the direction vector $$\mathbf{d} = (2,-3,2)$$. Hence their dot product is zero: $$(1+2t_0)(2) + (1-3t_0)(-3) + (1+2t_0)(2) = 0.$$

Simplify: $$2(1+2t_0) - 3(1-3t_0) + 2(1+2t_0) = 0$$ $$\Rightarrow 2 + 4t_0 - 3 + 9t_0 + 2 + 4t_0 = 0$$ $$\Rightarrow 1 + 17t_0 = 0$$ $$\Rightarrow t_0 = -\frac{1}{17}.$$

Substitute $$t_0$$ back to get the coordinates of $$P$$: $$a = 1 + 2\left(-\frac{1}{17}\right) = 1 - \frac{2}{17} = \frac{15}{17},$$ $$b = 1 - 3\left(-\frac{1}{17}\right) = 1 + \frac{3}{17} = \frac{20}{17},$$ $$c = 1 + 2\left(-\frac{1}{17}\right) = \frac{15}{17}.$$

Now compute $$34(a+b+c)$$: $$a+b+c = \frac{15}{17} + \frac{20}{17} + \frac{15}{17} = \frac{50}{17},$$ $$34(a+b+c) = 34 \times \frac{50}{17} = 2 \times 50 = 100.$$

Therefore,

$$34(a + b + c) = 100$$

Option C: 100