If $$\vec{x}$$ and $$\vec{y}$$ be two non-zero vectors such that $$|\vec{x} + \vec{y}| = |\vec{x}|$$ and $$2\vec{x} + \lambda\vec{y}$$ is perpendicular to $$\vec{y}$$, then the value of $$\lambda$$ is_____.

Let us translate the given geometric conditions into algebra. We are told that the two non-zero vectors $$\vec{x}$$ and $$\vec{y}$$ satisfy the length (magnitude) equation

$$|\vec{x}+\vec{y}| \;=\;|\vec{x}|.$$

First we square both sides, because the square of a magnitude is easier to expand by the dot product. Using the fact that, for any vector $$\vec{a},$$ we have $$|\vec{a}|^{2}=\vec{a}\cdot\vec{a},$$ we write

$$|\vec{x}+\vec{y}|^{2} \;=\; |\vec{x}|^{2}.$$

Now expand the left-hand side by the distributive property of the dot product:

$$ \begin{aligned} |\vec{x}+\vec{y}|^{2} &= (\vec{x}+\vec{y})\cdot(\vec{x}+\vec{y}) \\ &= \vec{x}\cdot\vec{x} + 2\vec{x}\cdot\vec{y} + \vec{y}\cdot\vec{y} \\ &= |\vec{x}|^{2} + 2\vec{x}\cdot\vec{y} + |\vec{y}|^{2}. \end{aligned} $$

Equating this with the right-hand side $$|\vec{x}|^{2}$$ gives

$$|\vec{x}|^{2} + 2\vec{x}\cdot\vec{y} + |\vec{y}|^{2} \;=\; |\vec{x}|^{2}.$$

The $$|\vec{x}|^{2}$$ terms on both sides cancel, leaving

$$2\vec{x}\cdot\vec{y} + |\vec{y}|^{2} \;=\;0.$$

We can solve this relation for the scalar product $$\vec{x}\cdot\vec{y}:$$

$$2\vec{x}\cdot\vec{y} \;=\; -|\vec{y}|^{2} \quad\Longrightarrow\quad \vec{x}\cdot\vec{y} \;=\; -\dfrac{|\vec{y}|^{2}}{2}.$$

Next, we use the perpendicularity condition that the vector $$2\vec{x}+\lambda\vec{y}$$ is perpendicular to $$\vec{y}.$$ Two vectors are perpendicular precisely when their dot product is zero. Therefore,

$$(2\vec{x}+\lambda\vec{y})\cdot\vec{y} \;=\;0.$$

Expand this dot product:

$$ \begin{aligned} (2\vec{x}+\lambda\vec{y})\cdot\vec{y} &= 2\vec{x}\cdot\vec{y} + \lambda\,\vec{y}\cdot\vec{y} \\ &= 2\vec{x}\cdot\vec{y} + \lambda|\vec{y}|^{2}. \end{aligned} $$

Set this equal to zero as required:

$$2\vec{x}\cdot\vec{y} + \lambda|\vec{y}|^{2} \;=\;0.$$

We already found $$2\vec{x}\cdot\vec{y} = -|\vec{y}|^{2}.$$ Substituting this value, we obtain

$$-|\vec{y}|^{2} + \lambda|\vec{y}|^{2} \;=\;0.$$

Factor out the common non-zero factor $$|\vec{y}|^{2}:$$

$$|\vec{y}|^{2}\,(\lambda - 1) \;=\;0.$$

Since $$\vec{y}$$ is non-zero, its magnitude $$|\vec{y}|^{2}$$ is strictly positive, so the only way for the product to vanish is

$$\lambda - 1 \;=\;0 \quad\Longrightarrow\quad \lambda = 1.$$

Hence, the correct answer is Option 1.