1. Given vectors

$\vec a=4\hat i+3\hat j=(4,3,0),\qquad \vec b=3\hat i-4\hat j+5\hat k=(3,-4,5)$

Let $\vec c=(x,y,z).$

We use the three given conditions to determine $\vec c$.

2. Condition involving $\vec a\times \vec b$

First compute the cross product:

Now, $\vec c\cdot(\vec a\times \vec b)+25=0$

gives $\vec c\cdot(15,-20,-25)=-25.$

So, $15x-20y-25z=-25$ $3x-4y-5z=-5 \qquad ...(1)$

3. Condition $\vec c\cdot(\hat i+\hat j+\hat k)=4$

$x+y+z=4 \qquad ...(2)$

4. Projection of $\vec c$ on $\vec a$ is 1

Scalar projection of $\vec c$ on $\vec a$ is

$\frac{\vec c\cdot \vec a}{|\vec a|}=1.$

Now, $|\vec a|=\sqrt{4^2+3^2}=5.$

Hence $\frac{4x+3y}{5}=1$ $4x+3y=5 \qquad ...(3)$

5. Solve for $x,y,z$

From (2): $z=4-x-y.$

Substitute into (1):

$3x-4y-5(4-x-y)=-5$ $3x-4y-20+5x+5y=-5$ $8x+y=15 \qquad ...(4)$

Now solve (3) and (4):

$4x+3y=5 \qquad ...(3)$ $8x+y=15 \qquad ...(4)$

From (4): $y=15-8x.$

Put into (3): $4x+3(15-8x)=5$ $4x+45-24x=5$ $-20x=-40$ $x=2.$

Then $y=15-16=-1.$

And from (2): $z=4-2-(-1)=3.$

Thus, $\vec c=(2,-1,3).$

6. Projection of $\vec c$ on $\vec b$

Scalar projection of $\vec c$ on $\vec b$ is

$\frac{\vec c\cdot \vec b}{|\vec b|}.$

Compute:

$\vec c\cdot \vec b=(2,-1,3)\cdot(3,-4,5)=2\cdot 3+(-1)(-4)+3\cdot 5=6+4+15=25.$

Also, $|\vec b|=\sqrt{3^2+(-4)^2+5^2}=\sqrt{9+16+25}=\sqrt{50}=5\sqrt2.$

Therefore, $\text{projection of }\vec c\text{ on }\vec b=\frac{25}{5\sqrt2}=\frac{5}{\sqrt2}.$

7. Option check

$\frac{5}{\sqrt2}$ corresponds to Option D.

8. Comparison with stored answer

Stored correct answer: D

Our derived answer: D

So the derived answer agrees with the stored answer.