Foot of Perpendicular from Origin to Line Intersecting Vectors

To find the equation of the line $L$, we first need to determine its direction vector $\vec{d}$. Since the line $L$ is perpendicular to both $\vec{v}_1 = 2\hat{i} + 2\hat{j} + \hat{k}$ and $\vec{v}_2 = \hat{i} + 2\hat{j} + 2\hat{k}$, its direction vector is given by the cross product of $\vec{v}_1$ and $\vec{v}_2$:

$\vec{d} = \vec{v}_1 \times \vec{v}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 2 & 1 \\ 1 & 2 & 2 \end{vmatrix}$

Expanding the determinant, we get: $\vec{d} = \hat{i}(4 - 2) - \hat{j}(4 - 1) + \hat{k}(4 - 2) = 2\hat{i} - 3\hat{j} + 2\hat{k}$

The line $L$ passes through the point $(1, 1, 1)$ with direction ratios $(2, -3, 2)$. Thus, any general point $P(a, b, c)$ on the line $L$ can be expressed in terms of a parameter $t$ as: $P(a, b, c) = (1 + 2t, 1 - 3t, 1 + 2t)$

Since $P(a, b, c)$ is the foot of the perpendicular from the origin $O(0, 0, 0)$ to line $L$, the position vector $\vec{OP}$ must be perpendicular to the direction vector $\vec{d}$ of the line $L$. Therefore: $\vec{OP} \cdot \vec{d} = 0$

Substituting the components: $(1 + 2t)(2) + (1 - 3t)(-3) + (1 + 2t)(2) = 0$ $2 + 4t - 3 + 9t + 2 + 4t = 0$ $17t + 1 = 0 \implies t = -\frac{1}{17}$

Now, we calculate the sum $a + b + c$: $a + b + c = (1 + 2t) + (1 - 3t) + (1 + 2t) = 3 + t$

Substituting $t = -\frac{1}{17}$: $a + b + c = 3 - \frac{1}{17} = \frac{50}{17}$

We are required to find the value of $34(a + b + c)$: $34(a + b + c) = 34 \times \frac{50}{17} = 2 \times 50 = 100$

Hence, the correct option is C.