If the projection of the vector $$\hat{i} + 2\hat{j} + \hat{k}$$ on the sum of the two vectors $$2\hat{i} + 4\hat{j} - 5\hat{k}$$ and $$-\lambda\hat{i} + 2\hat{j} + 3\hat{k}$$ is 1, then $$\lambda$$ is equal to _________

First, we identify all the vectors involved. The vector whose projection we are talking about is $$\vec{A}=\,\hat{i}+2\hat{j}+\hat{k}.$$ The two vectors that must be added are $$\vec{B}=\,2\hat{i}+4\hat{j}-5\hat{k}$$ and $$\vec{C}= -\lambda\hat{i}+2\hat{j}+3\hat{k}.$$

We form their sum: $$\vec{D}= \vec{B}+\vec{C} =\bigl(2-\lambda\bigr)\hat{i} +\bigl(4+2\bigr)\hat{j} +\bigl(-5+3\bigr)\hat{k} =(2-\lambda)\hat{i}+6\hat{j}-2\hat{k}.$$

The question tells us that the scalar projection (sometimes called the component) of $$\vec{A}$$ on $$\vec{D}$$ equals $$1$$. We state the standard formula: For any two vectors $$\vec{u}$$ and $$\vec{v}$$, the scalar projection of $$\vec{u}$$ on $$\vec{v}$$ is $$\text{proj}_{\vec{v}}\vec{u}= \dfrac{\vec{u}\cdot\vec{v}}{|\vec{v}|}\,.$$

Applying this formula, we must have $$\dfrac{\vec{A}\cdot\vec{D}}{|\vec{D}|}=1.$$

Now we compute the dot product in the numerator. Writing both vectors in component form, $$\vec{A}=(1,\,2,\,1),\qquad \vec{D}=(2-\lambda,\,6,\,-2).$$ Hence $$\vec{A}\cdot\vec{D}=1\,(2-\lambda)+2\,(6)+1\,(-2) =(2-\lambda)+12-2 =12-\lambda.$$

Next we need the magnitude of $$\vec{D}$$. Using the formula $$|\vec{v}|=\sqrt{v_x^2+v_y^2+v_z^2},$$ we have $$|\vec{D}|=\sqrt{(2-\lambda)^2+6^2+(-2)^2} =\sqrt{(2-\lambda)^2+36+4} =\sqrt{(2-\lambda)^2+40}.$$ Therefore the projection condition becomes $$\dfrac{12-\lambda}{\sqrt{(2-\lambda)^2+40}}=1.$$

We clear the square root by squaring both sides: $$(12-\lambda)^2=(2-\lambda)^2+40.$$

Now we expand each square completely. On the left, $$(12-\lambda)^2=\lambda^2-24\lambda+144.$$ On the right, $$(2-\lambda)^2+40=(\lambda-2)^2+40 =\bigl(\lambda^2-4\lambda+4\bigr)+40 =\lambda^2-4\lambda+44.$$

Equating the two expressions and cancelling the identical $$\lambda^2$$ terms on both sides, we get $$\lambda^2-24\lambda+144=\lambda^2-4\lambda+44 \;\Longrightarrow\; -24\lambda+144=-4\lambda+44.$$

Bringing all terms to one side, $$-24\lambda+144+4\lambda-44=0 \;\Longrightarrow\; -20\lambda+100=0.$$

Dividing by $$-20$$, we find $$\lambda = 5.$$

So, the answer is $$5$$.