Let `C:vecr(t)=x(t)hati+y(t)hatj+z(t)hatk` be a differentiable curve i.e., `exists lim_(hto0) (vecr(t+h)-vecr(t))/(h) forall t`
`therefore vecr'(t)=x'(t)hati+y'(t)hatj+z'(t)hatk`
`vecr'(t)` is tangent to the curve `C` at the point `P[x(t),y(t),z(t)], vecr'(t)` points in the direction of increasing `t`.
The point `P` on the curve `vecr(t)=(1-2t)hati+t^(2)hatj+2e^(2(t-1))hatk` at which the tangent vector `vecr'(t)` is parallel to the radius vector `vecr(t)` is:

Options

A. (a) `(-1,1,2)`
B. (b) `(1,-1,2)`
C. (c) `(-1,1,-2)`
D. (d) `(1,1,2)`

Explanation

To solve the problem, we need to find the point \( P \) on the curve defined by the vector function \( \vec{r}(t) = (1 - 2t) \hat{i} + t^2 \hat{j} + 2e^{2(t-1)} \hat{k} \) where the tangent vector \( \vec{r}'(t) \) is parallel to the radius vector \( \vec{r}(t) \).

### Step 1: Find the derivative \( \vec{r}'(t) \)

The first step is to differentiate \( \vec{r}(t) \) with respect to \( t \):

\[ \vec{r}'(t) = \frac{d}{dt}[(1 - 2t) \hat{i} + t^2 \hat{j} + 2e^{2(t-1)} \hat{k}] \]

Calculating the derivatives:

- For \( x(t) = 1 - 2t \):
\[ x'(t) = -2 \]

- For \( y(t) = t^2 \):
\[ y'(t) = 2t \]

- For \( z(t) = 2e^{2(t-1)} \):
\[ z'(t) = 2 \cdot 2e^{2(t-1)} = 4e^{2(t-1)} \]

Thus, we have:

\[ \vec{r}'(t) = -2 \hat{i} + 2t \hat{j} + 4e^{2(t-1)} \hat{k} \]

### Step 2: Set up the condition for parallel vectors

The tangent vector \( \vec{r}'(t) \) is parallel to the radius vector \( \vec{r}(t) \) if:

\[ \frac{x'(t)}{x(t)} = \frac{y'(t)}{y(t)} = \frac{z'(t)}{z(t)} \]

Substituting the expressions we found:

- \( x(t) = 1 - 2t \)
- \( y(t) = t^2 \)
- \( z(t) = 2e^{2(t-1)} \)

We can write:

\[ \frac{-2}{1 - 2t} = \frac{2t}{t^2} = \frac{4e^{2(t-1)}}{2e^{2(t-1)}} \]

### Step 3: Simplify and solve the equations

From the first two ratios:

\[ \frac{-2}{1 - 2t} = \frac{2t}{t^2} \]

Cross-multiplying gives:

\[ -2t^2 = 2t(1 - 2t) \]

This simplifies to:

\[ -2t^2 = 2t - 4t^2 \]
\[ 2t^2 + 2t = 0 \]
\[ 2t(t + 1) = 0 \]

Thus, \( t = 0 \) or \( t = -1 \).

Now, from the second and third ratios:

\[ \frac{2t}{t^2} = 2 \]

Cross-multiplying gives:

\[ 2t = 2t^2 \]
\[ t^2 - t = 0 \]
\[ t(t - 1) = 0 \]

Thus, \( t = 0 \) or \( t = 1 \).

### Step 4: Evaluate at \( t = 1 \)

We need to check which of these values gives us a point on the curve:

1. For \( t = 0 \):
\[ \vec{r}(0) = (1 - 2 \cdot 0) \hat{i} + 0^2 \hat{j} + 2e^{2(0-1)} \hat{k} = 1 \hat{i} + 0 \hat{j} + 2e^{-2} \hat{k} \]

2. For \( t = 1 \):
\[ \vec{r}(1) = (1 - 2 \cdot 1) \hat{i} + 1^2 \hat{j} + 2e^{2(1-1)} \hat{k} = -1 \hat{i} + 1 \hat{j} + 2 \hat{k} \]

Thus, the point \( P \) is:

\[ P = (-1, 1, 2) \]

### Conclusion

The point \( P \) on the curve where the tangent vector is parallel to the radius vector is:

\[ \boxed{(-1, 1, 2)} \]