1. Given line

The line is

$$\frac{x+3}{-3}=\frac{y-1}{1}=\frac{z-5}{5}.$$

Its point-direction form can be read as:

• A point on the line: $P(-3,1,5)$
• Direction vector: $\vec d_1 = (-3,1,5)$

So the given line is

$$\vec r = (-3,1,5) + \lambda(-3,1,5).$$

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2. Condition for two lines to be coplanar

Two lines are coplanar if the scalar triple product of:

• direction vector of first line,
• direction vector of second line,
• vector joining one point on first line to one point on second line,

is zero.

That is,

$$[\vec d_1,\vec d_2,\overrightarrow{P_1P_2}] = 0.$$

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3. Check each option

We take $P_1=(-3,1,5)$ and $\vec d_1=(-3,1,5)$.

Option A

$$\frac{x+1}{-1}=\frac{y-2}{2}=\frac{z-5}{4}$$

Point: $Q_A=(-1,2,5)$

Direction vector:

$$\vec d_A = (-1,2,4)$$

Vector joining points:

$$\overrightarrow{P_1Q_A} = (-1+3,2-1,5-5)=(2,1,0)$$

Now compute scalar triple product:

$$\begin{vmatrix} -3 & 1 & 5 \\ -1 & 2 & 4 \\ 2 & 1 & 0 \end{vmatrix}$$

Expanding:

$$= -3\begin{vmatrix}2 & 4 \\ 1 & 0\end{vmatrix} -1\begin{vmatrix}-1 & 4 \\ 2 & 0\end{vmatrix} +5\begin{vmatrix}-1 & 2 \\ 2 & 1\end{vmatrix}$$ $$= -3(2\cdot 0 - 4\cdot 1) - ( -1\cdot 0 - 4\cdot 2 ) + 5((-1)(1)-2\cdot 2)$$ $$= -3(-4) - (-8) + 5(-1-4) =12+8-25=-5 \neq 0$$

So A is not coplanar.

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Option B

$$\frac{x+1}{-1}=\frac{y-2}{2}=\frac{z-5}{5}$$

Point: $Q_B=(-1,2,5)$

Direction vector:

$$\vec d_B = (-1,2,5)$$

Vector joining points:

$$\overrightarrow{P_1Q_B}=(2,1,0)$$

Scalar triple product:

$$\begin{vmatrix} -3 & 1 & 5 \\ -1 & 2 & 5 \\ 2 & 1 & 0 \end{vmatrix}$$

Expanding:

$$= -3\begin{vmatrix}2 & 5 \\ 1 & 0\end{vmatrix} -1\begin{vmatrix}-1 & 5 \\ 2 & 0\end{vmatrix} +5\begin{vmatrix}-1 & 2 \\ 2 & 1\end{vmatrix}$$ $$= -3(2\cdot 0-5\cdot 1) -((-1)\cdot 0-5\cdot 2)+5((-1)(1)-2\cdot 2)$$ $$= -3(-5)-(-10)+5(-5)=15+10-25=0$$

So B is coplanar.

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Option C

$$\frac{x-1}{-1}=\frac{y-2}{2}=\frac{z-5}{5}$$

Point: $Q_C=(1,2,5)$

Direction vector:

$$\vec d_C=(-1,2,5)$$

Vector joining points:

$$\overrightarrow{P_1Q_C}=(1+3,2-1,5-5)=(4,1,0)$$

Scalar triple product:

$$\begin{vmatrix} -3 & 1 & 5 \\ -1 & 2 & 5 \\ 4 & 1 & 0 \end{vmatrix}$$

Expanding:

$$= -3\begin{vmatrix}2 & 5 \\ 1 & 0\end{vmatrix} -1\begin{vmatrix}-1 & 5 \\ 4 & 0\end{vmatrix} +5\begin{vmatrix}-1 & 2 \\ 4 & 1\end{vmatrix}$$ $$= -3(-5)-((-1)\cdot 0-5\cdot 4)+5((-1)(1)-2\cdot 4)$$ $$=15-(-20)+5(-9)=15+20-45=-10\neq 0$$

So C is not coplanar.

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Option D

$$\frac{x+1}{1}=\frac{y-2}{2}=\frac{z-5}{5}$$

Point: $Q_D=(-1,2,5)$

Direction vector:

$$\vec d_D=(1,2,5)$$

Vector joining points:

$$\overrightarrow{P_1Q_D}=(2,1,0)$$

Scalar triple product:

$$\begin{vmatrix} -3 & 1 & 5 \\ 1 & 2 & 5 \\ 2 & 1 & 0 \end{vmatrix}$$

Expanding:

$$= -3\begin{vmatrix}2 & 5 \\ 1 & 0\end{vmatrix} -1\begin{vmatrix}1 & 5 \\ 2 & 0\end{vmatrix} +5\begin{vmatrix}1 & 2 \\ 2 & 1\end{vmatrix}$$ $$= -3(-5) - (1\cdot 0-5\cdot 2)+5(1\cdot 1-2\cdot 2)$$ $$=15-(-10)+5(-3)=15+10-15=10\neq 0$$

So D is not coplanar.

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4. Conclusion

Only Option B satisfies the coplanarity condition.

Therefore, the required line is

$$\boxed{\frac{x+1}{-1}=\frac{y-2}{2}=\frac{z-5}{5}}$$