$$\begin{gathered} A\mathrm{line}\mathrm{through}(1,1,1)\mathrm{and}\mathrm{perpendicular}\mathrm{to\; both } \\ \hat{i}+2\hat{j}+2\hat{k},2\hat{i}+2\hat{j}+\hat{k} \\ \mathrm{ let}(a,b,c)\mathrm{be}\mathrm{foot}\mathrm{of}\mathrm{perpendicular}\mathrm{from}\mathrm{origin} \\ \mathrm{then}34(a+b+c)\mathrm{is} \end{gathered}$$

The correct answer is 100.

Step-by-step Explanation:

Let the given vectors perpendicular to the line be:
$\overrightarrow{v_1}=\hat{i}+2\hat{j}+2\hat{k}$
and
$\overrightarrow{v_2}=2\hat{i}+2\hat{j}+\hat{k}$

The direction vector $\overrightarrow{d}$ of the line is perpendicular to both $\overrightarrow{v_1}$ and $\overrightarrow{v_2}$. Therefore, we can find $\overrightarrow{d}$ by taking the cross product of $\overrightarrow{v_1}$ and $\overrightarrow{v_2}$:

Expanding the determinant:
$\overrightarrow{d}=\hat{i}(2·1-2·2)-\hat{j}(1·1-2·2)+\hat{k}(1·2-2·2)$
$\overrightarrow{d}=\hat{i}(2-4)-\hat{j}(1-4)+\hat{k}(2-4)$
$\overrightarrow{d}=-2\hat{i}+3\hat{j}-2\hat{k}$

Since the line passes through the point $(1,1,1)$, any general point $(a,b,c)$ on this line can be expressed in terms of a real parameter $t$ as:
$a=1-2t$
$b=1+3t$
$c=1-2t$

We are given that $(a,b,c)$ is the foot of the perpendicular from the origin $(0,0,0)$ to the line. Therefore, the vector from the origin to this point, $a\hat{i}+b\hat{j}+c\hat{k}$, must be perpendicular to the direction vector of the line:
$(a\hat{i}+b\hat{j}+c\hat{k})·(-2\hat{i}+3\hat{j}-2\hat{k})=0$
$-2a+3b-2c=0$

Substitute the expressions for $a$, $b$, and $c$ in terms of $t$ into this equation:
$-2(1-2t)+3(1+3t)-2(1-2t)=0$
$-2+4t+3+9t-2+4t=0$
$17t-1=0$
$t=\frac{1}{17}$

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

Substituting $t=\frac{1}{17}$:
$a+b+c=3-\frac{1}{17}=\frac{51-1}{17}=\frac{50}{17}$

Finally, find the value of $34(a+b+c)$:
$34(a+b+c)=34·\frac{50}{17}=2·50=100$